A ball is thrown up at 8m.s from the roof which is 60m high.it strikes the balcony below after 4s.the ball then bounces off the balcony and strikes the ground.ignore the effects of friction.calculate magnitude of the velocity at which the ball strike the balcony?
To calculate the magnitude of the velocity at which the ball strikes the balcony, we can use the laws of motion and the concept of conservation of energy.
First, let's break down the problem:
1. Initial velocity: The ball is thrown up with an initial velocity of 8 m/s.
2. Initial height: The ball is thrown from a roof that is 60m high.
3. Time: The ball strikes the balcony after 4 seconds.
4. Final height: We need to determine the final height at which the ball strikes the balcony.
To find the final height, we can use the formula for calculating displacement in vertical motion:
s = ut + (1/2)gt^2
where:
s = displacement (final height - initial height)
u = initial velocity (8 m/s)
t = time (4 seconds)
g = acceleration due to gravity (-9.8 m/s^2, considering downward direction)
Plugging in the given values:
s = (8 * 4) + (0.5 * -9.8 * (4^2))
Simplifying the equation:
s = 32 - 78.4
s = -46.4 m
The negative sign indicates that the final height is below the initial height.
Now, we can calculate the magnitude of the velocity at which the ball strikes the balcony using the principle of conservation of energy.
The initial kinetic energy of the ball (K.E.) is equal to its potential energy (P.E.) at the final height.
K.E. = P.E.
(1/2)mv^2 = mgh
where:
m = mass of the ball (mass cancels out in the equation)
v = velocity at which the ball strikes the balcony
g = acceleration due to gravity
h = final height
Rearranging the equation to solve for v:
v = √(2gh)
Plugging in the values:
v = √(2 * 9.8 * 46.4)
v ≈ √(908.16)
v ≈ 30.13 m/s
Therefore, the magnitude of the velocity at which the ball strikes the balcony is approximately 30.13 m/s.