Question

A ball is projected upwards from the top of tower with a velocity 50 ms^–1 making angle 30^o 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

Answer: (C)

6.33 sec

Answer 2

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 $t = \frac{u}{g}\left\lbrack 1 + \sqrt{1 + \frac{2gh}{u^{2}}} \right\rbrack$

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

Initial vertical component of velocity $u\sin\theta = 50\sin 30 = 25m/s.$

∴ $t = \frac{25}{9.8}\left\lbrack 1 + \sqrt{1 + \frac{2 \times 9.8 \times 70}{(25)^{2}}} \right\rbrack$ = 6.33 sec.