Question

A ball is projected upwards from the top of tower with a velocity $50m{s}^{-1}$ making angle ${30}^{\circ }$ with the horizontal. The height of the tower is 70 m. After how many seconds from the instant of throwing will the ball reach the ground

• A. 2.33 sec
• B. 5.33 sec
• C. 6.33 sec
• D. 9.33 sec

Answer 1

The correct option is C

6.33 sec

Formula for calculation of time to reach the body on the ground from the tower of height ‘h’ (If it is thrown vertically up with velocity u) is given by $t=\frac{u}{g}[1+\sqrt{1+\frac{2gh}{u^2}}]$

So we can resolve the given velocity in vertical direction and can apply the above formula.

Initial vertical component of velocity $usin\theta =50sin30=25m/s$

$\therefore t=\frac{25}{9.8}[1+\sqrt{1+\frac{2\times 9.8\times 70}{(25{)}^2}}]=7.11sec$