Question

A ball is thrown from upwards from the top of a a velocity of $20m/s$ .The ball strikes the earth after $5s$ of its throwing. What is the height of the tower? What will be the velocity of the ball while striking the earth? $(g=9.8m/s^2)$

Answer 1

By using laws of motion we get

$S=-ut+\frac{1}{2}a{t}^2$

$u=20\frac{m}{\sec }$

By substituting the time and initial velocity in above equation and acceleration due ot gravity as$9.8\frac{m}{{\sec }^2}$

$S=-20\times \left(5\right)+\frac{1}{2}\times 9.8\times {\left(5\right)}^2$

$S=22.5m$

Again by using

$v=u+at$

The velocity of the body is calculated as

$v=-20+9.8\times 5$

$v=29\frac{m}{\sec }$

Then velocity of the ball while stinking is$29\frac{m}{\sec }$