Question

A ball is thrown from a point $1m$ above the ground. The initial velocity is $20m/s$ at an angle of $40$ degrees above the horizontal. Calculate the speed of the ball at the highest point in the trajectory. (in $m/s$)

• A. $15.3$
• B. $16.3$
• C. $14.2$
• D. $12$

Answer 1

The correct option is A $15.3$

Given : $u=20m/s$ $\theta ={40}^o$

Initially, horizontal component of velocity $u_x=u\cos \theta$

$⟹u_x=20\times \cos 40=20\times 0.766=15.32m/s$

Since there is no acceleration of the ball along the horizontal direction, so its velocity component along the horizontal component remains the same all the time.

At highest point, vertical component of velocity is zero.

Thus final velocity of ball is equal to $u_x$.

$v=u_x=15.32m/s$