Question

An object is launched directly upward with an initial velocity of 64 feet per second from a platform 80 feet high. What will be the object's maximum height? When will it attain that height? When will the object reach ground level?

Answer 1

Answer:

a) 63.6 ft

b) 1.99 seconds

c) 4.98 seconds

Step-by-step explanation:

a) The object will reach maximum height when its final velocity, v, is 0 m/s.

We apply one of Newton's equations of motion to solve this:

$v^2 = u^2 - 2gh$

where u = initial velocity = 64 ft/s

g = acceleration due to gravity = 32.2 ft/s

h = height reached by object.

Note: the equation has a negative sign because the object is going upwards, against gravity.

Hence:

$0^2 = 64^2 - 2*32.2*h\\\\=> 64.4h = 4096\\\\h = 4096 / 64.4\\\\h = 63.6 ft$

The object is now 63.6 ft above the platform (143.6 ft above the ground).

b) Time taken to get to that height can be gotten by using another one of Newton's equations:

$v = u - gt$

=> $0 = 64 - 32.2t$

$32.2t = 64\\\\=> t = 64 / 32.2 \\\\t = 1.99 secs$

It took the object 1.99 secs to get to that height.

c) When the object begins falling to ground level, it is at a height, h =  143.6 ft and begins moving at initial velocity, u = 0 m/s.

Applying another of Newton's equations of motion, we have that:

$h = ut + \frac{1}{2}gt^2$

where t is the time taken to hit the floor after it begins its descent.

Therefore,

$143.6 = 0*t + \frac{1}{2}*32.2t^2\\ \\143.6 = 16.1t^2\\\\t^2 = 143.6 / 16.1\\\\t^2 = 8.92\\\\t = \sqrt{8.92} \\\\t = 2.99 secs$

This is the time it will take the object to hit the floor after it begins its descent.

Therefore, the total time it will take for the object to hit the ground after it is launched will be:

T = time taken to reach maximum height + time taken to hit the floor after it begins descending

T = 1.99 + 2.99

T = 4.98 seconds