Question

A man is sitting on the shore of a river. He is in the line of 1.0 m long boat and is 5.5 m away from the centre of the boat. He wishes to throw an apple into the boat. If he can throw the apple only with a speed of 10 m/s, find the minimum and maximum angles of projection for successful shot. Assume that the point of projection and the edge of the boat are in the same horizontal level.

Answer 1

Given:
Length of the boat = 1.0 m
Distance between the man and the centre of the boat (R) = 5.5 m
Initial speed (u) of throwing the apple by the man = 10 m/s
Acceleration due to gravity (g) = 10 m/s²
We know that the horizontal range is given by
$$\begin{gathered} R=\frac{u^2\sin 2\mathrm{\alpha }}{g} \\ \Rightarrow 5=\frac{{\left(10\right)}^2\sin 2\mathrm{\alpha }}{10} \\ \Rightarrow \sin 2\mathrm{\alpha }=\frac{1}{2} \\ \Rightarrow \alpha =15°\mathrm{or}75° \end{gathered}$$


Similarly, for the end point of the boat, i.e., point C, we have:
Horizontal range (R) = 6 m
$$\begin{gathered} R=\frac{u^2\sin 2\mathrm{\alpha }}{g} \\ \Rightarrow 6=\frac{{\left(10\right)}^2\sin 2\mathrm{\alpha }}{10} \\ \Rightarrow \sin 2\mathrm{\alpha }=\frac{3}{5} \\ \Rightarrow \alpha =18°\mathrm{or}71° \end{gathered}$$
For a successful shot, the angle of projection α with initial speed 10 m/s may vary from 15° to 18° or from 71° to 75°. The minimum angle is 15° and the maximum angle is 75°, but there is an interval of 53° for which the successful shot is not allowed. We can show this by putting the successive value of α from 15° to 75°.