Question

A stone is dropped from a height of 9 meters above the ground. If the height functions can be modelled by the equation h(t) = a ─ $ {{\text{t}}^{2}} $ , where t is the time in seconds, h is height in meters, and a is the initial height, how many seconds does it take for the stone to hit the ground?

Answer 1

Hint: The height h of the stone the distance of the stone from the ground, whereas a is the height from which the stone is dropped. It is given that the motion is governed by the equation h(t) = a ─ $ {{\text{t}}^{2}} $ . We will substitute the height h at the ground level into the equation of motion h(t) = a ─ $ {{\text{t}}^{2}} $ and substitute the given initial height. Then we will solve the equation to find the value of t and thus we can find the time required for the stone to reach the ground from a height of 9 meters.

Complete step-by-step answer:
The equation of motion is given as h(t) = a ─ $ {{\text{t}}^{2}} $ .
We can see from the equation that as the t, that is the time increases, the value of h, that is the height at that instant of time decreases, whereas a is the initial height and remains constant.
It is given to us that the stone is dropped from a height of 9 meters. Thus, a = 9.
So, we substitute a = 9 in the equation h(t) = a ─ $ {{\text{t}}^{2}} $ .
 $ \Rightarrow $ h(t) = 9 ─ $ {{\text{t}}^{2}} $
Now, we will find the instantaneous height at t = 0.
 $ \Rightarrow $ h(0) = 9 ─ $ {{\left( 0 \right)}^{2}} $
 $ \Rightarrow $ h(0) = 9
Now, if the stone reaches the ground, its instantaneous height will be 0.
Thus, substitute h(t) = 0 in h(t) = 9 ─ $ {{\text{t}}^{2}} $
 $ \Rightarrow $ 0 = 9 ─ $ {{\text{t}}^{2}} $
 $ \Rightarrow $ $ {{\text{t}}^{2}} $ = 9
 $ \Rightarrow $ t = 3 or –3
But t cannot be negative, thus t = 3.
Thus, it takes 3 seconds for the stone to reach the ground.

Note: It is important to check equations of motion with t = 0, so that we can keep in mind any anomalies, if present, in the motion. At t=0 we get the initial position of the object before the displacement.