Question

A ball is thrown from the ground with a velocity of $20\sqrt{3}{ms}^{-1}$ making an angle of ${60}^o$ with the horizontal. The ball will be at a height of $40m$ from the ground after a time $t$ equal to $(g=10{ms}^{-2})$

• A. $\sqrt{2}s$
• B. $\sqrt{3}s$
• C. $2s$
• D. $3s$

Answer 1

The correct option is C $2s$

Initial velocity $u=20\sqrt{3}m/s$
Angle $\theta =60$
height, h=40 m
h=$u\sin \theta t-\frac{g{t}^2}{2}$
$40=20\sqrt{3}\sin 60t-5t^2$
$5t^2-30t+40=0$
$t^2-6t+8=0$
t=2 & 6 seconds
The ball will attain 40 meters in 2 seconds while ascending and 6 seconds after it is projected while descending.