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A ball is thrown straight up in the air with an initial velocity of 50 feet per second and an initial height of 5 feet. How high will the ball be after 3 seconds?
Step 1
Write down the information you already know:
v1= 50 feet per second [up]
a= 9.81 m/s2 [down]= -9.81 m/s2 [up] = -32.185 feet/s2 [down] (1 metre= 3.28083989501312 feet)
t= 3s
d= ?
Step 2
Solve:
d= v1t+(1/2)at2
d= 50(3)+(1/2)(-32.185)(3)2
d= 150-144.8325
d= 5.1675 feet
*Remember you need to add the initial height
d=5+5.1675
d= 10.1675 feet= 10 feet
What else can I help you with?
How to find the height of the projectile launch if the angle velocity and the distance are given?
Get the value of initial velocity. Get the angle of projection. Break initial velocity into components along x and y axis. Apply the equation of motion .
If a cricket jumps off the ground with an initial vertical velocity of 4 ft per second what is an equation the height in ft of the crisket as a function of time in seconds since it jumped?
4ft*Ns=H
Will a ball drop rest reach the ground quicker than the one lunched from the same height but with and initial horizontal velocity?
No. What counts in this case is the vertical component of the velocity, and the initial vertical velocity is zero, one way or another.
How many feet does a balloon move each second after how many seconds will a balloon be at height of 40 feet?
The answer depends on what gas the balloon contains, its initial velocity and the forces - gravity, buoyancy, cross-wind - acting on it.
If the launch angle is 15 degrees and the initial velocity is 50.0 meters per second what is the range?
If the initial velocity is 50 meters per second and the launch angle is 15 degrees what is the maximum height? Explain.
An arrow is shot straight up at an initial velocity of 200 meters per seconds. How long will it be before beginning to fall?
The arrow will begin to fall when its velocity becomes negative, which will happen after it reaches its maximum height and starts to descend. The time it takes for the arrow to reach its peak height can be calculated using the formula: time = (final velocity - initial velocity) / acceleration. After reaching the peak, the arrow will take the same amount of time to fall back down.
An object projected up with initial velocity v attains a height h Another object projected up with initial velocity 2v will attain what height?
The height attained by an object projected up is directly proportional to the square of its initial velocity. So, if an object with initial velocity v attains a height h, then an object with initial velocity 2v will attain a height of 4 times h.
What is the equation to calculate the maximum height of an object given an initial velocity and initial launch height?
height=acceletation(t^2) + velocity(t) + initial height take (T final - T initial) /2 and place it in for time and there you go
What is A football kicked off at an angle 20 from the ground reaches a maximum height of 4.7 m What is the initial velocity of the kick?
To find the initial velocity of the kick, you can use the equation for projectile motion. The maximum height reached by the football is related to the initial vertical velocity component. By using trigonometric functions, you can determine the initial vertical velocity component and then calculate the initial velocity of the kick.
How can you throw a projectile so that it has zero speed at the top?
To have zero speed at the top, you need to throw the projectile with an initial velocity such that it reaches its maximum height at that point. This requires the initial velocity to be exactly equal to the velocity that would be attained due to gravity when the projectile falls from that height. The angle of projection should be such that the vertical component of the initial velocity cancels out the velocity due to gravity.
A ball thrown vertically upward returns to the starting point in 8 seconds.find its initial velocity.?
The total time of flight for a ball thrown vertically upwards and returning to its starting point is twice the time taken to reach maximum height. Therefore, the time taken to reach maximum height is 4 seconds. Given that the acceleration due to gravity is -9.8 m/s^2, using the kinematic equation v = u + at, where v is the final velocity (0 m/s at maximum height), u is the initial velocity, a is the acceleration due to gravity, and t is the time, you can solve for the initial velocity. Substituting the values, u = 9.8 * 4 = 39.2 m/s. Therefore, the initial velocity of the ball thrown vertically upward is 39.2 m/s.
How can one determine the maximum height reached by an object launched with a given initial velocity?
To determine the maximum height reached by an object launched with a given initial velocity, you can use the formula for projectile motion. The maximum height is reached when the vertical velocity of the object becomes zero. This can be calculated using the equation: Maximum height (initial velocity squared) / (2 acceleration due to gravity) By plugging in the values of the initial velocity and the acceleration due to gravity (which is approximately 9.81 m/s2 on Earth), you can find the maximum height reached by the object.
A diver jumps from a 48ft cliff into the ocean with an initial velocity of 8 seconds when does he hit the water?
Assuming the acceleration due to gravity is -32 ft/s², we can use the kinematic equation h = (1/2)gt² + v₀t + h₀, where h is the final height (0 ft), g is the acceleration due to gravity, v₀ is the initial velocity (8 ft/s), and h₀ is the initial height (48 ft) to solve for t. Plug in the values and solve for t to find the time it takes for the diver to hit the water.
How do initial velocity affect range and height of a projectile?
Increasing the initial velocity of a projectile will increase both its range and height. Higher initial velocity means the projectile will travel further before hitting the ground, resulting in greater range. Additionally, the increased speed helps the projectile reach a higher peak height before it begins to descend back down.
If at the Earth's surface a projectile is launched straight up at a speed of 9 km per sec to what height will it rise Ignore air resistance and the rotation of the Earth?
9 km/s = 9000m/s Gravity decreases the velocity of the object by 9.8 m/s each second. The velocity at the top is 0 m/s Equation 1: Velocity final = velocity initial - (9.8 m/s × time) Final velocity =0 m/s Initial velocity = 9000m/s 0 = 9000 - 9.8 t 9.8 t = 9000 t = 9000÷ 9.8 t = 918 seconds Average velocity = (9000 + 0 ) ÷ 2 =4500 m/s Height = average velocity × time Height = 4500 m/s ×918 seconds=4,131,000 meters = 4,131 Km. If you do not want to round, this equation will find the answer more accurately. (velocity final) 2 - (velocity initial) 2 = 2 × acceleration × distance m/s2 0 - 90002 = 2 × 9.8 × d d = 4,132,653.061 meters = 4,132.653061Km I do not know of any measuring tool that measures that precisely!
What relationship exists between the initial velocity and the maximum height reached by an object thrown upward?
Ignoring air resistance, I get this formula:Maximum height of a vertically-launched object = 1.5 square of initial speed/GI could be wrong. In that case, the unused portion of my fee will be cheerfully refunded.
How does the height from which an object was dropped affect its average velocity?
The height from which an object is dropped does not affect its average velocity. Average velocity depends on the overall displacement and time taken to achieve that displacement, regardless of the initial height of the object.
