From the top of a tower 19.6 ,m high, a ball | vedclass

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Question

(A) Given that, the height of the tower is $h = 19.6 \,m$.The angle of the line joining the point of projection to the point on the ground with the ho

Vedclass · Accepted Answer

$9.8$

Vedclass · Answer

(A) Given that, the height of the tower is $h = 19.6 \,m$.The angle of the line joining the point of projection to the point on the ground with the horizontal is $	heta = 45^{\circ}$.In the right-angled triangle formed by the height $h$ and horizontal range $R$, we have:$	an 	heta = \frac{h}{R}$$	an 45^{\circ} = \frac{19.6}{R}$$1 = \frac{19.6}{R} \Rightarrow R = 19.6 \,m$.The time taken $t$ to reach the ground is given by:$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 	imes 19.6}{9.8}} = \sqrt{4} = 2 \,s$.The horizontal distance $R$ is given by:$R = u 	imes t$$19.6 = u 	imes 2$$u = \frac{19.6}{2} = 9.8 \,m/s$.Thus, the initial velocity of the ball is $9.8 \,m/s$.

From the top of a tower $19.6 \,m$ high, a ball is thrown horizontally. If the line joining the point of projection to the point where it hits the ground makes an angle of $45^{\circ}$ with the horizontal, then the initial velocity of the ball is (in $\,m/s$)

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