To find the final velocity of the ball when it hits the ground, we can use the formula ( v = u + at ), where ( u = 5 , \text{m/s} ) (initial velocity), ( a = -9.81 , \text{m/s}^2 ) (acceleration due to gravity), and ( t = 5.5 , \text{s} ). Plugging in the values, ( v = 5 + (-9.81)(5.5) \approx -49.955 , \text{m/s} ) (the negative sign indicates downward direction). To find the height from which the ball was thrown, we can use the formula ( h = ut + \frac{1}{2}at^2 ), which gives ( h = 5(5.5) + \frac{1}{2}(-9.81)(5.5^2) \approx -113.0 , \text{m} ) (indicating it fell below the initial height).
Velocity final = vi + at = 49 m/s displacement = vi * t + ½2at² = 122.5 m vi = 0 a ≈ 9.8 t = 5
At any time 't' seconds after the ball is released,until it hits the ground,h = 5 + 48 t - 16.1 t2
4ft*Ns=H
Acceleration = (change in velocity) / (time for the change)9.8 = (change in velocity) / (2 seconds)9.8 x 2 = change in velocity = 19.6 meters per second .Hint: The mass of the object and the height of the building are there just tothrow you off balance. You don't need either of them to answer the question.
To determine how long it takes for the cricket to land back on the ground after jumping with an initial vertical velocity of 4 ft per second, we can use the formula for the time of flight in projectile motion. The time to reach the maximum height is given by ( t = \frac{v}{g} ), where ( v ) is the initial velocity and ( g ) is the acceleration due to gravity (approximately 32 ft/s²). In this case, it takes ( t = \frac{4}{32} = 0.125 ) seconds to reach the peak. Since the time to ascend and descend is equal, the total time until the cricket lands back on the ground is ( 2 \times 0.125 = 0.25 ) seconds.
The velocity of the rock as it reaches the ground after 3.5 seconds of free fall can be calculated using the equation v = gt, where v is the final velocity, g is the acceleration due to gravity (approximately 9.81 m/s^2), and t is the time in seconds. Substituting the values, v = 9.81 m/s^2 * 3.5 s = 34.335 m/s. So, the velocity of the rock as it reaches the ground is approximately 34.34 m/s.
Velocity final = vi + at = 49 m/s displacement = vi * t + ½2at² = 122.5 m vi = 0 a ≈ 9.8 t = 5
Terminal velocity is typically reached within 10-12 seconds when falling from a height, depending on factors such as air resistance and the height of the fall.
If it was thrown horizontally or dropped, and hit the ground 3.03 seconds later, then it hit the ground moving at a speed of 29.694 meters (97.42-ft) per second. If it was tossed at any angle not horizontal, and hit the ground 3.03 seconds later, we need to know the direction it was launched, in order to calculate the speed with which it hit the ground.
At any time 't' seconds after the ball is released,until it hits the ground,h = 5 + 48 t - 16.1 t2
4ft*Ns=H
The height of the bridge can be calculated using the formula: distance = 0.5 * acceleration due to gravity * time^2. Given that the rock takes 8 seconds to hit the water, the height of the bridge would be approximately 313.6 meters.
A suspension bridge is supported from a height of 30 feet. The length of the wire used to suspend the bridge, from the ground to the top, is 65 feet. What is the angle of elevation from the base of the bridge to the point of suspension?
To answer this question one would need to know the rock's initial height and velocity.
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
It will fall with increasing velocity due to gravity and reach the peak velocity just before hitting the ground.
Assuming the stone was dropped from rest, we can calculate the height of the bridge using the kinematic equation: h = 0.5 * g * t^2, where h is the height of the bridge, g is the acceleration due to gravity (9.8 m/s^2), and t is the time of fall (2.1 seconds). Plugging in the values, we get h = 0.5 * 9.8 * (2.1)^2 = 22.33 meters. Therefore, the height of the bridge is approximately 22.33 meters.