Question

A stone is thrown vertically upward with a speed of $49m{s}^{-1}$. Find the velocity of the stone one second before it reaches the maximum height.

• A. $4.9m{s}^{-1}$
• B. $9.8m{s}^{-1}$
• C. $13.6m{s}^{-1}$
• D. $19.6m{s}^{-1}$

Answer 1

Answer: (B)

$9.8m{s}^{-1}$

Initial velocity of the stone is $u=49m/s$

Acceleration: $a=g=9.8m{s}^{-2}$

While ascending, the stone will deccelerate and finally its velocity will become zero at the maximum height.

Let $t$ is the time taken by the stone to reach the maximum height.

Using the equation $v=u-gt$

$t=\frac{v-u}{-g}=\frac{0-49}{-9.8}=5sec$

We need to find the velocity of the stone at $t=5-1=4sec$

Again, Using the equation $v=u-gt$

$v=49-(9.8\times 4)=49-39.2=9.8m/s$