From the top of a tower $19.6 \mathrm{~m}$ 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

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Aicharya · Accepted Answer

Given that, height of tower, $h=19.6 \mathrm{~m}$

Angle of line joining from point of projection to the point of ground with horizontal, $ \begin{aligned} \theta & =45^{\circ} \\ \text { In } \triangle A O B, \tan \theta & =\frac{h}{R} \\ \tan 45^{\circ} & =\frac{h}{R} \Rightarrow 1=\frac{h}{R} \\ \Rightarrow \quad R & =h=19.6 \mathrm{~m} \end{aligned} $ where, $R$ be the horizontal distance. We know that, time taken to reach from point $A$ to $B$, $ t=\sqrt{\frac{2 h}{g}} $ By substituting the values, we get $ t=\sqrt{\frac{2 \times 19.6}{9.8}}=2 \mathrm{~s} $ Now, Horizontal distance $=$ Horizontal velocity $\times$ Time $ \begin{aligned} & \Rightarrow \quad R=u t \\ & 19.6=u \times 2 \\ & \end{aligned} $ [By substituting the values of $R$ and $t$, we get] $\therefore$ Initial velocity, $u=9.8 \mathrm{~m} / \mathrm{s}$

From the top of a tower $19.6 \mathrm{~m}$ high, a ball is thrown horizontally. If the line joining the…

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