Question

a stone is thrown straight up from the edge of a roof, 950 feet above the ground, at a speed of 16 feet per second. a. remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 2 seconds later? b. at what time does the stone hit the ground? c. what is the velocity of the stone when it hits the ground?

Answer 1

a) The stone is 1046 feet above the ground

b) It takes 1 second to return to the ground

c) The speed as it returns to the ground is 259 feet/s.

What is motion under gravity?

We know that when an object is thrown up or thrown down, the motion of the object would occur under the influence of gravity and we have to apply the equations of kinematics in each case.

Given that;

a) h = ut - 1/2gt^2

h = height attained

u = initial velocity

g = acceleration due to gravity

t = time taken

h = (16 *2) - 1/2(-32 * (2)^2)

h = 32 + 64

h = 96 feet

The distance above the ground = 96 feet + 950 feet = 1046 feet

b) v = u + gt

Given that v = 0 at the maximum height

0 = 16 + (-32) * t

t = 16/32

t = 0.5 secs

To go up and return;

2(0.5 secs) = 1 second

c) v^2 = u^2 + 2gh

v = √(16)^2 + (2 * 32 * 1046)

v = 259 feet/s

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