Question

A boy kicks his football (soccer ball) with an initial speed of 30 m/s at an angle 45o
above the horizontal aiming to hit a 20 m-tall post which is 25 m away from him.
Neglecting air resistance, find: a) with what velocity will he hit the post, b) where
exactly (x, y) will he hit it?

Answer 1

Answer on Question #40952 - Physics - Mechanics | Kinematics | Dynamics

A boy kicks his football (soccer ball) with an initial speed of $30\mathrm{m/s}$ at an angle 45° above the horizontal aiming to hit a $20\mathrm{m}$-tall post which is $25\mathrm{m}$ away from him. Neglecting air resistance, find: a) with what velocity will he hit the post, b) where exactly $(x, y)$ will he hit it?

Solution:

$\mathrm{L} = 25 \mathrm{~m}$ – horizontal range of the ball;

$\mathrm{V} = 30 \frac{\mathrm{m}}{\mathrm{s}}$ – speed of the soccer ball;

$\alpha = 45{}^{\circ}$ – the angle of projection;

$\mathrm{H} = 20 \mathrm{~m}$ – height of the post;

$\mathrm{h}$ – height at which the ball will hit the post;

Equation of the motion for the ball, directed at angle $\alpha$: (t – time of the flight)

$\mathrm{V_x} = \mathrm{V}\cos \alpha ;\mathrm{V_y} = \mathrm{V}\sin \alpha ;$

$$\mathrm{x: L = Vt \cos \alpha}$$ $$t = \frac {L}{V \cos \alpha}$$ $$y: h = Vt \sin \alpha - \frac {g t ^ {2}}{2}$$

(2)in(1)

$$\begin{array}{l} h = \frac {L}{V \cos \alpha} \cdot V \sin \alpha - \frac {g \left(\frac {L}{V \cos \alpha}\right) ^ {2}}{2} = L \tan \alpha - \frac {g L ^ {2}}{2 V ^ {2} \cos^ {2} \alpha} = \\ = 25 \mathrm{m} \cdot \tan 45{}^{\circ} - \frac {9.8 \frac {\mathrm{m}}{\mathrm{s}^{2}} \cdot (25 \mathrm{m})^{2}}{2 \cdot \left(30 \frac {\mathrm{m}}{\mathrm{s}}\right)^{2} \cdot \cos 45{}^{\circ}} = 18.2 \mathrm{m} \\ \end{array}$$

Ball hit the post, so it traveled distance $L$ along the X-axis and distance $h$ along Y-axis, hence, ball will hit the post in point $(L, h) = (25m, 18.2m)$

Principle of conservation of energy: initial energy of the ball is equal to the energy when the ball hit the post:

$$W_1 = W_2 \quad (3)$$

Initial energy of the ball (initial level on the ground, thus $PE = 0$)

$$W_1 = W_{KE1} = \frac{mV^2}{2} \quad (4)$$

Moment of hitting the post:

$$W_2 = W_{KE2} + W_{PE2} = \frac{mV_1^2}{2} + mgh \quad (5)$$

(5) and (4) in (3):

$$\frac{mV^2}{2} = \frac{mV_1^2}{2} + mgh$$ $$V_1^2 = V^2 - 2gh$$ $$V_1 = \sqrt{V^2 - 2gh} = \sqrt{\left(30\frac{m}{s}\right)^2 - 2 \cdot 9.8\frac{N}{kg} \cdot 18.2m} = 23.3\frac{m}{s}$$

Answer: a) ball will hit the post with velocity $23.3\frac{m}{s}$

b) point of the hit: $(25m, 18.2m)$.