Question

a ball is projected from ground level with a speed of 28 m s⁻¹ at an angle of 30° to the horizontal as shown below. there is a wall of height h at a distance of 16 m from the point of projection of the ball. air resistance is negligible.
(a) calculate the initial magnitudes of (i) the horizontal velocity of the ball, (ii) the vertical velocity of the ball.
a ball is projected from ground level with a speed of 28 m s⁻¹ at an angle of 30° to the horizontal as shown below. there is a wall of height h at a distance of 16 m from the point of projection of the ball. air resistance is negligible.
(b) the ball just passes over the wall. determine the maximum height of the wall.

Explanation:

Step1: Find initial horizontal velocity

Use $v_{x0}=v_0\cos\theta$. Given $v_0 = 28\ m/s$ and $\theta = 30^{\circ}$, so $v_{x0}=28\cos30^{\circ}=28\times\frac{\sqrt{3}}{2}=14\sqrt{3}\ m/s\approx24.2\ m/s$.

Step2: Find initial vertical velocity

Use $v_{y0}=v_0\sin\theta$. Given $v_0 = 28\ m/s$ and $\theta = 30^{\circ}$, so $v_{y0}=28\sin30^{\circ}=28\times\frac{1}{2}=14\ m/s$.

Step3: Find time to reach the wall

In horizontal direction, $x = v_{x0}t$. Since $x = 16\ m$ and $v_{x0}=14\sqrt{3}\ m/s$, then $t=\frac{x}{v_{x0}}=\frac{16}{14\sqrt{3}}=\frac{8}{7\sqrt{3}}\ s$.

Step4: Find height of the wall

In vertical direction, use $y = v_{y0}t-\frac{1}{2}gt^{2}$. Substitute $v_{y0}=14\ m/s$, $t = \frac{8}{7\sqrt{3}}\ s$ and $g = 9.8\ m/s^{2}$.
\[

$$\begin{align*} y&=14\times\frac{8}{7\sqrt{3}}-\frac{1}{2}\times9.8\times(\frac{8}{7\sqrt{3}})^2\\ &=\frac{16}{\sqrt{3}}-\frac{4.9\times64}{49\times3}\\ &=\frac{16}{\sqrt{3}}-\frac{64}{30}\\ &=\frac{16}{\sqrt{3}}-\frac{32}{15}\\ &\approx\frac{16\times1.732}{3}-\frac{32}{15}\\ &=\frac{27.712}{3}-\frac{32}{15}\\ &=\frac{138.56 - 32}{15}\\ &=\frac{106.56}{15}\approx7.1\ m \end{align*}$$

\]

Answer:

(a)(i) $14\sqrt{3}\ m/s\approx24.2\ m/s$
(a)(ii) $14\ m/s$
(b) Approximately $7.1\ m$