Question

A ball is thrown from a balcony with an initial speed of 31 m/s and at a projection angle of 24 degrees. The point of release is 8.2 m above flat terrain. (a) What is the horizontal distance from the release point to where the ball hits the ground? (b) What is the straight-line distance from the release point to where the ball hits the ground?

Answer 1

First we have to find the x and y component of the initial velocity.

x component of initial velocity:-

$\sf u_x = u \:cos\theta \\ \longrightarrow \sf u_x = 31 \times cos(24^o) \\ \longrightarrow \sf u_x = 31 \times 0.914 \\ \longrightarrow \sf u_x = 28.334\:m/s$

y component of initial velocity:-

$\sf u_y = u\:sin\theta \\ \longrightarrow \sf u_y = 31 \times sin(24^o) \\ \longrightarrow \sf u_y = 31 \times 0.407 \\ \longrightarrow \sf u_y = 12.617$

In y component:-

$\sf s_y = u_yt + \dfrac{1}{2}a_yt^2 \\ \longrightarrow \sf -H = 12.617t + \dfrac{1}{2}(-g)t^2 \\ \longrightarrow \sf -8.2 = 12.617t - 5t^2$

On solving the above quadratic equation for 't', we get,

$\sf t_1 = -0.536 \\ \sf t_2 = 3.059$

As time cannot be negative,

$\sf t = t_1 = 3.059 \:s$

In x component:-

$\sf s_x = u_xt + \dfrac{1}{2}a_xt^2$

Time will be same in both x and y component.

$\longrightarrow \sf R = (28.334 \times 3.059) + \dfrac{1}{2}0(3.059)^2$

$\longrightarrow \underline{\underline{\sf{\green{R = 86.673\:m}}}}$

As the angle between the tower/balcony and the ground is $\sf 90^o,$

$\sf d = \sqrt{H^2 + R^2} \\ \longrightarrow \sf d = \sqrt{(8.2)^2 + (86.673)^2} \\ \longrightarrow \sf d = \sqrt{67.24 + 7512.209} \\ \longrightarrow \sf d = \sqrt{7579.449}$

$\longrightarrow \underline{\underline{\sf{\green{d = 87.060\:m}}}}$

Note that I have approximated the values, so the answer which is given may slightly differ by a decimal or so.