Answer to Question #138969 in Physics for lucas

Question

A football is kicked at a velocity of 15 m/s at an angle of 25 o to the horizontal. Calculate:

a) The total flight time

b) The maximum height

c) The range of the ball.

Answer 1

a) Let's write the equation of motion of the ball in horizontal and vertical directions:

$$x=v_0tcos\theta, (1)$$ $$y=v_0tsin\theta-\dfrac{1}{2}gt^2, (2)$$

here, $x$ is the horizontal displacement of the ball (or the range of the ball), $v_0 = 15 \ \dfrac{m}{s}$ is the initial velocity of the ball, $t$ is the total flight time of the ball, $\theta=25^{\circ}$ is the launch angle, $y$ is the vertical displacement of the ball (or the height) and $g = 9.8 \ \dfrac{m}{s^2}$ is the acceleration due to gravity.

We can find the total flight time of the ball from the second equation (since ball returns to the ground, $y = 0$):

$$0=v_0tsin\theta-\dfrac{1}{2}gt^2,$$ $$t = \dfrac{2v_0sin\theta}{g}=\dfrac{2\cdot 15\ \dfrac{m}{s}\cdot sin25^{\circ}}{9.8 \ \dfrac{m}{s^2}}=1.29 \ s.$$

b) Let's first find the time that ball takes to reach the maximum height from the kinematic equation:

$$v_y = v_0sin\theta - gt_{rise},$$ $$0=v_0sin\theta - gt_{rise},$$ $$t_{rise} = \dfrac{v_0sin\theta}{g}.$$

Then, we can substitute $t_{rise}$into the second equation and find the maximum height:

$$y_{max}=v_0sin\theta\cdot\dfrac{v_0sin\theta}{g}-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2,$$ $$y_{max}=\dfrac{v_0^2sin^2\theta}{2g} = \dfrac{(15\ \dfrac{m}{s})^2 \cdot sin^2 25^{\circ}}{2\cdot 9.8 \ \dfrac{m}{s^2}} = 2.05 \ m.$$

c) Finally we can substitute the total flight time into the first equation and find the range of the ball:

$$x=v_0cos\theta \dfrac{2v_0sin\theta}{g} = \dfrac{v_0^2sin2\theta}{g},$$ $$x=\dfrac{(15\ \dfrac{m}{s})^2 \cdot sin2\cdot 25^{\circ}}{9.8 \ \dfrac{m}{s^2}}=17.59\ m.$$

Answer:

a) $t =1.29 \ s.$

b) $y_{max}= 2.05 \ m.$

c) $x=17.59\ m.$