Q2.4 Projectile motion (S)

A ball is thrown horizontally with speed u from a height of 20 m. It hits the ground with a speed 3u. Find the value of u, and the angle at which the ball strikes the ground.
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Hint

Start with Denote the launch height by h.

You’ll need to deal separately with the horizontal and vertical components of the motion.

 

Solution

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Solution

Denote the launch height by h. The initial horizontal and vertical components of the velocity are
\[ v_{x0} =u \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} v_{y0} =0 \]
The (constant) acceleration has components
\[ a_{x} =0 \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} a_{y} =-g \]
Graphic - No title - balltraj
On impact, then (appealing to Key Point 1.4)
\[ v_x = v_{x0} +a_xt = v_{x0} = u \hspace{0.5cm}\mbox{\rm and}\hspace{0.5cm} v_y^2 = v_{y0}^2 +2a_y(y-y_0) = -2g \times -h =2gh \]
The speed of impact (3u) is the magnitude of the final velocity. Thus
\[ \left[3u\right]^2 = [v]^2 = v_x^2 +v_y^2 = u^2 + 2gh \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} 8u^2 = 2gh \]
Then (setting g=9.8 ms-2)
\[ u= \sqrt{\frac{gh}{4}} = 7.0\,ms^{-1} (2sf) \]
The angle the final velocity makes with the vertical is φ where (see the sketch; and note that it is the magnitude of vy we need to insert in this equation)
\[ tan {\phi} = \frac{v_x}{\mid{v_y}\mid} = \frac{u}{\sqrt{8}u} =\frac{1}{\sqrt{8}} \]
implying φ=19 (2sf).