A ball is thrown upward with an initial velocity of 96 feet per second from the top of a 100 foot building. (a) When well the ball attain its maximum height? (b) When will the ball hit the ground? Give a decimal answers correct to 2 decimal places
(a) When g*t = 96 ft/s
t = 96/32.2 = ___ s
(b) When 100 + 96 t - (32.2/2)*t^2 = 0
Use the quadratic equation and take the positive root.
To solve this problem, we can use the equation of motion for an object in freefall:
s = ut + (1/2)at^2
Where:
s = displacement (height in this case)
u = initial velocity (96 ft/s)
a = acceleration due to gravity (-32 ft/s^2, negative due to the direction)
t = time
(a) To find when the ball will attain its maximum height, we need to determine the time it takes for the ball to reach its highest point. At this point, the ball's velocity becomes zero.
Using the formula for final velocity:
v = u + at
0 = 96 - 32t
Solving for t:
32t = 96
t = 96/32
t = 3 seconds
So, it takes 3 seconds for the ball to reach its maximum height.
(b) To find when the ball will hit the ground, we need to determine the time it takes for the ball to travel back down from its maximum height to the ground.
Using the formula for displacement:
s = ut + (1/2)at^2
We know the initial displacement (s) is 100 feet (height of the building), and the initial velocity (u) is 0 at the highest point. We also know the acceleration (a) is -32 ft/s^2.
Substituting the values:
100 = 0*t + (1/2)*(-32)*t^2
100 = -16t^2
Dividing both sides by -16:
t^2 = -100/-16
t^2 = 6.25
Taking the square root of both sides:
t = √6.25
t ≈ 2.50 seconds
So, it takes approximately 2.50 seconds for the ball to hit the ground.
Therefore, the answers are:
(a) The ball will attain its maximum height in 3 seconds.
(b) The ball will hit the ground in approximately 2.50 seconds.