Question

A stone is thrown straight up with an initial speed of 12.0m/s, off the edge of a 75m tall building. A. How long does it take the stone to reach the ground?

B. How high above the ground does the stone go?
C. At what time is the stone 50.0m from the bottom of the building?
D. How fast is the stone moving when it is 50m from the bottom of the building?

Answer 1

A) 3.11 seconds

B) 82.33 m

C) 5.32 seconds

D) 25.2 m/s

Step-by-step explanation:

t = Time taken

u = Initial velocity = 12 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

`v=u+at\\\Rightarrow 0=12-9.8* t\\\Rightarrow (-12)/(-9.81)=t\\\Rightarrow t=1.22 \s`

Time taken to reach maximum height is 1.22 seconds

`s=ut+(1)/(2)at^2\\\Rightarrow s=12* 1.22+(1)/(2)* -9.81* 1.22^2\\\Rightarrow s=7.33\ m`

So, the stone would travel 7.33 m up

B) So, total height stone would fall is 7.33+75 = 82.33 m

`s=ut+(1)/(2)at^2\\\Rightarrow 82.33=0t+(1)/(2)* 9.81* t^2\\\Rightarrow t=\sqrt{(82.33* 2)/(9.81)}\\\Rightarrow t=4.1\ s`

A) Total time taken by the stone to reach the ground from the time it left the person's hand is 4.1+1.22 = 5.32 seconds.

C) When s = 82.33-50 = 33.33 m

`s=ut+(1)/(2)at^2\\\Rightarrow 32.33=0t+(1)/(2)* 9.81* t^2\\\Rightarrow t=\sqrt{(32.33* 2)/(9.81)}\\\Rightarrow t=2.57\ s`

The time it takes from the person's hand throwing the ball to the distance of 50 m from the ground is 2.57+1.22 = 3.777 seconds

D)

`v=u+at\\\Rightarrow v=0+9.81* 2.57\\\Rightarrow v=25.2\ m/s`

The stone is moving at 25.2 m/s when it is 50m from the bottom of the building