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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
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B
5s
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C
7s
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D
9s
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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 between
point $$B$$ 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 $$T = T_1 + T_2 = 5 + 2 = 7 \sec$$

1166699_1657_ans_68f25105ff5d494aae645df99d2784e0.png

Physics

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