Question

A stone is thrown vertically upward with a speed of 28 m/s. (a) Find the maximum height reached by the stone. (b) Find its velocity one second before it reaches the maximum height. (c) Does the answer of part (b) change if the initial speed is more than 28 m/s such as 40 m/s or 80 m/s?

Answer 1

Given:
Initial velocity with which the stone is thrown vertically upwards, u = 28 m/s
When the stone reaches the ground, its final velocity (v) is 0.
Also,
a = g = −9.8 m/s² (Acceleration due to gravity)
(a) Maximum height can be found using the equation of motion.
Thus, we have:
v² − u² = 2as
$\mathrm{s}=\frac{v^2-u^2}{2a}$
On putting respective values, we get:
$s=\frac{0^2-{28}^2}{2\left(-9.8\right)}=40\mathrm{m}$

(b) Total time taken by the stone to reach the maximum height:
$t=\frac{\left(v-u\right)}{a}$
$\Rightarrow t=\frac{\left(0-28\right)}{-9.8}=2.85\mathrm{s}$
As per the question, we need to find the velocity of the stone one second before it reaches the maximum height.
t' = 2.85 − 1 = 1.85 s
Again, using the equation of motion, we get:
v' = u + at' = 28 − 9.8 × 1.85
⇒ v' = 28 − 18.13 = 9.87 m/s
Hence, the velocity is 9.87 m/s

(c) No answer of part (b) will not change, as after one second, the velocity becomes zero for any initial velocity and acceleration (a = − 9.8 m/s²) remains the same. For any initial velocity more than 28 m/s, only the maximum height increases.