0
What else can I help you with?
How is it that A ball is projected straight up from the ground with a speed of 7.61 meters per second To what maximum height will the ball go when gravity equals 9.80 meters per second squared?
Initial upward speed = 7.61 m/sFinal upward speed (at the point of maximum height) = 0Time to reach maximum height = (7.61) / (9.8) = 0.77653 secondAverage speed during that time = 1/2 ( 7.61 + 0) = 3.805 m/sHeight = 3.805 x 0.77653 = 2.9547 meters (rounded) = about 9.7 feetDoesn't seem like much of a height for a strong toss; but the math looks OK.
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.
How high does a ball go if it is thrown up with an initial speed of 10 meters per second?
applying the equation: (VxV) -(UXU) = 2aS on reaching the maximum height the ball stops so the final velocity(V) becomes =0. initial velocity as given is (U)=10 m/s. ------------------------------------------------------------------------------------2 acceleration is here gravity in downward direction so (a)=-9.81 m/s so applying the above stated formula the height reached by the ball(S)=5.0968 m (approx)
A long jumper leaves the ground at an angle 20 to horizontal and at speed of 11ms.how far does he jump and what the maximum height reached?
Making the improbable assumption that the jumper experiences no air resistance, he will jump 3.97 metres, and reach a height of 0.72 metres.
When throwing a ball straight up when in the air with an initial velocity of 10 meters per second what will be it maximum height?
'Maximum height' means the exact point at which the velocity changes from upward to downward. At that exact point, the magnitude of the velocity is zero. It doesn't matter what the velocity was when it left your hand. That number determines the maximum height, but the velocity at that height is always zero. --------------------------------------------------------- Thus using the formula: (vf)e2 = (vi)e2+2*a*d vf = final velocity = 0 m/s vi = initial velocity = 10 m/s a = acceleration = gravity = - 9.81 m/s/s d = displacement (distance) = ? e is designating that the next figure is an exponent in the formula So the formula is: (0)e2 = (10)e2 + (2 * -9.81 * d) 0 = 100 + -19.62d adding 19.62d to both sides of the equation 19.62d = 100 dividing by 19.62 d = ~ 5.097 meters
What is the maximum height reached by a projectile shot straight up from the Earth's surface at a speed of 1.10104 km/hr?
The maximum height reached by a projectile shot straight up from the Earth's surface at a speed of 1.10104 km/hr is determined by the formula: Maximum height (initial velocity)2 / (2 acceleration due to gravity) Given that the initial velocity is 1.10104 km/hr, we can convert this to m/s by multiplying by 1000/3600. The acceleration due to gravity is approximately 9.81 m/s2. Plugging in the values, we can calculate the maximum height reached by the projectile.
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.
How can one determine the maximum height reached by a projectile?
To determine the maximum height reached by a projectile, you can use the formula: maximum height (initial vertical velocity)2 / (2 acceleration due to gravity). This formula calculates the height based on the initial vertical velocity of the projectile and the acceleration due to gravity.
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.
What is the maximum height reached by a projectile shot straight up from the Earth's surface at a speed of 10000 km/hr?
The maximum height reached by a projectile shot straight up from the Earth's surface at a speed of 10000 km/hr is approximately 138.9 kilometers.
What is the maximum height reached by a projectile shot straight up from the Earth's surface at a speed of 8000 km/hr?
The maximum height reached by a projectile shot straight up from the Earth's surface at a speed of 8000 km/hr is approximately 222.22 kilometers.
What is the maximum height reached by a projectile shot straight up from the Earth's surface at a speed of 9000 km/hr?
The maximum height reached by a projectile shot straight up from the Earth's surface at a speed of 9000 km/hr is approximately 225 kilometers.
When a ball is throw upward how does the height reached by the ball depend on its initial speed?
The height reached by a ball thrown upward depends on its initial speed: the higher the initial speed, the higher the maximum height reached. This is because a greater initial speed gives the ball more kinetic energy, allowing it to overcome gravity and reach a higher position before gravity brings it back down.
What is the maximum height reached by a projectile shot straight up from the Earth's surface at a speed of 1.20104 km/hr?
The maximum height reached by a projectile shot straight up from the Earth's surface at a speed of 1.20 x 104 km/hr is approximately 72 kilometers.
How can one determine the maximum height reached in projectile motion?
To determine the maximum height reached in projectile motion, you can use the formula: textMaximum height left(fracv02 sin2(theta)2gright) where ( v0 ) is the initial velocity, ( theta ) is the launch angle, and ( g ) is the acceleration due to gravity. By plugging in these values, you can calculate the maximum height the projectile reaches.
A boy kicks a ball vertically upwards with an initial velocity of 12m/s. Calculate the time taken by the ball to reach the maximum hight and the maximum, height reached by the ball?
The time taken by the ball to reach the maximum height is 1 second. The maximum height reached by the ball is 36 meters.
What is the maximum height reached by a projectile shot straight up from the Earth's surface at a speed of 1.00 x 104 km/hr?
The maximum height reached by a projectile shot straight up from the Earth's surface at a speed of 1.00 x 104 km/hr is approximately 2.78 kilometers.
