Question

# A ball is projected upwards from the top of the tower with a velocity $$50$$ m/s making an angel $$30^{\circ}$$ 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
2s
B
5s
C
7s
D
9s

Solution

## The correct option is B $$7 s$$Given: initial velocity (v) $$= 50 \,m/s$$height of the tower (h) $$= 70 \,m$$angle with the horizontal $$(\theta) = 30^o$$we take $$g = 10 \,m/sec^2$$First ball goes from point $$A$$ to $$B$$.$$\therefore$$ for projectile motion $$AB$$ is$$T_1 = \dfrac{2 \,u \sin \theta}{g} = \dfrac{2 \times 50 \times \sin \,30^o}{10}$$$$T_1 = \dfrac{2 \times 50 \times \dfrac{1}{2}}{10}$$$$T_1 = 5 \sec$$Now time taken for covering distance betweenpoint $$B$$ to $$C$$ is $$T_2$$$$\Longrightarrow h = u \,T_2 + \dfrac{1}{2} g \,T^2_2$$      $$70 = u \sin 30^o \,T_2 + \dfrac{1}{2} \times 10 T_2^2$$      $$70 = 50 \times \dfrac{1}{2} T_2 + 5 T^2_2$$      $$70 = 25 \,T_2 + 5 T^2_2$$$$\Longrightarrow 5T^2_2 + 25 \,T_2 - 70 = 0$$$$\Longrightarrow T^2_2 + 5 \,T_2 - 14 = 0$$$$\Longrightarrow (T_2 - 2) (T_2 + 7) = 0$$$$\Longrightarrow T_2 = 2$$ and $$T_2 = -7$$ (not possible)$$\therefore$$ Total time $$T = T_1 + T_2 = 5 + 2 = 7 \sec$$Physics

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