Question

1. a football player kicks a stationary ball of mass 0.45 kg towards a wall. the initial speed of the ball after the kick is 19 m s⁻¹ and the ball does not rotate. air resistance is negligible and there is no wind. diagram not to scale (a.i) the ball leaves the ground at an angle of 22°. the horizontal distance from the initial position of the edge of the ball to the wall is 11 m. calculate the time taken for the ball to reach the wall. 1 (a.ii) the top of the wall is 2.4 m above the ground. deduce whether the ball will hit the wall. 3

Explanation:

Step1: Find horizontal - velocity component

The initial velocity of the ball is $v_0 = 19\ m/s$ and the angle of projection is $\theta=22^{\circ}$. The horizontal - velocity component is given by $v_{0x}=v_0\cos\theta$. So, $v_{0x}=19\cos22^{\circ}\approx19\times0.9272 = 17.62\ m/s$.

Step2: Calculate time to reach the wall

We know that in horizontal motion (where there is no acceleration, $a_x = 0$), the horizontal displacement $x = v_{0x}t$. Given $x = 11\ m$ and $v_{0x}\approx17.62\ m/s$, we can solve for $t$. Rearranging the formula $t=\frac{x}{v_{0x}}$, so $t=\frac{11}{17.62}\approx0.624\ s$.

Step3: Find vertical displacement at time $t$

The initial vertical - velocity component is $v_{0y}=v_0\sin\theta=19\sin22^{\circ}\approx19\times0.3746 = 7.12\ m/s$. The equation for vertical displacement is $y = v_{0y}t-\frac{1}{2}gt^{2}$, where $g = 9.8\ m/s^{2}$ and $t = 0.624\ s$.
Substitute the values: $y=7.12\times0.624-\frac{1}{2}\times9.8\times(0.624)^{2}$.
$y = 4.44 - 1.91=2.53\ m$.

Answer:

(a.i) $t\approx0.62\ s$
(a.ii) Since $y = 2.53\ m>2.4\ m$, the ball will not hit the wall.