A ball is dropped from the top of a tall building of height = 50 m. About how long does it take for the ball to hit the ground?
(Neglect air resistance)
b) If the ball has lost half the magnitude of its impact momentum immediately after it re-coils, to what height does the ball reach afterits first rebound?
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Let the acceleration due to gravity at place be g = 9.8 m/s^2.
The motion under gravitation of the ball dropped from the top of the guilding is given by:
s(t) = (1/2)gt^2, where s is the vertically downward distance traveeled by the ball from the top of the building in time t seconds.
S the time to reach the ground t is given by the equation when s(t) = 50m.
So, 50 = (1/2)gt^2. g= 9.8m/s assumed.
t^2 = 2*50/g
t = sqrt(100/9.8) = 3.1944 seconds approximately.
So the ball takes 3.1944 seconds.
To find the momentum of the ball when it hits the ground.
Momentum = mv , where m is the mass of the ball * v velocity of the ball when it hits the ground.......(1)
So the velocity of the ball when it strikes the ground = u +gt , where u = 0 the initial velocity when the ball was dropped.
v = gt = 9.8* sqrt(100/9.8) = 31.304951695
When the ball recoils it has only half the momentum . So the ball has an initial velocity of 31.3045/2 = 15.65247584 m/s.
So the again the ball follows the law of motion:
Smax = (v^2-u^2)/2g, where u = initial velocity = 15.6524...m/s as found above. v = final velocity when the ball has 0 velocity and starts returning to ground reaching the maximum height Smax.
So Smax = (0-15.65247584^2)/(2*-9.8) = 12.5 meter.
Therefore , the ball with half the momentum reach 12.5 meter of maximum height (that is quarter of the height of the building.)
A ball is dropped from the top of a tall building of height 50 m. The time it takes to reach the ground has to be determined with negligible air resistance.
Use the formula s = u*t + (1/2)*a*t^2 where s is the total distance traveled in time t, u is the initial velocity and a is the acceleration.
In this problem, s = 50 as the ball is at a height 50 m, u = 0 as the ball is dropped and a = 9.8, which is the gravitational acceleration. This gives:
50 = 0 + (1/2)*9.8*t^2
t = `sqrt (500/49) ~~ 3.19` seconds
When the ball strikes the ground, it loses half the magnitude of its impact momentum. Let the height it reaches when it rebounds be H.
The momentum of the ball on impact is m*v where m is the mass and v is the velocity with which it strikes the ground.
Here, v = 9.8*3.19 = 31.3 m/s. As half the momentum is lost, the velocity of the ball is 15.65 m/s. Its kinetic energy is (1/2)*m*v^2 = m*122.5. This is equal to m*g*H.
m*9.8*H = m*122.5
H = 122.5/9.8
The ball rebounds to a height of 12.5 m.
Any object will accellerate due to gravity -9.81 m/s^2. By integrating this acceleration over time, one obtains the velocity at any time t.
v = gt + v0. In this case, the ball started at rest, so the initial velocity v0 is zero.
By integrating velocity over time, one obtains the position y at any time t.
y = 1/2gt^2 + y0. We know that y0, the initial height of the ball, is 50 m.
So, when does the ball hit the ground (y = 0). Solve for t:
0 = 1/2 (-9.81 m/s^2) t^2 + 50 --> 100s^2 = 9.81 t^2 --> t = 3.2 s
You can answer the second question by examining the energy of the system. The kinetic energy when the ball hits the ground is determined by it's velocity. 1/2 mv^2. The velocity at t = 3.2s is v = g*3.2 = 10.2 m/s.
Uk = 52*m J/kg
The problem states that the ball looses half it's energy to the impact. So the new kinetic energy is 26*m J/kg. How high will it go? All of the kinetic energy will go to potential energy according to the equation Up = mgh. So,
Uk = Up --> 52m = mgh --> h = 52/g = 5.3 meters.
gravity is 9.14 meater per second so:
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