A ball is projected upwards from the top of tower with a velocity 50 ms−1 mak

Question

A ball is projected upwards from the top of tower with a velocity

50 m 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

Options:

A . 2.33 sec

B . 5.33 sec

C . 7.11 sec

D . 6.33 sec

Answer: Option C
:
C
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{2 g h}{u^{2}}}]

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

u s i n θ = 50 s i n 30 = 25 m / s

∴ t = \frac{25}{9.8} [1 + \sqrt{1 + \frac{2 \times 9.8 \times 70}{(25)^{2}}}] = 7.11 s e c

A Ball Is Projected Upwards From The Top Of Tower With A Vel...

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