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. Find the maximum height of the ball above the ground. (in $m$)

• A. $8.5$
• B. $9.5$
• C. $10$
• D. $11$

Answer 1

The correct option is B $9.5$

Initial height of the ball $h_i=1m$

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

The ball, during its projectile motion, reaches up to a further maximum height $H$, given by

$H=\frac{u^2{\sin }^2\theta }{2g}$

$⟹H=\frac{(20{)}^2\times {\sin }^2(40)}{2\times 9.8}=\frac{(20{)}^2\times (0.64{)}^2}{2\times 9.8}=8.5m$

Thus maximum height achieved by ball $H_{max}=1+8.5=9.5m$