Question

A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. What is the maximum height attained by the ball? Round the answer to the nearest tenth of a meter

Answer 1

Answer:

8.6 m

Explanation:

The motion of a soccer ball is a motion of a projectile, with a uniform motion along the horizontal (x-) direction and an accelerated motion along the vertical (y-) direction, with constant acceleration $a=g=-9.8 m/s^2$ towards the ground (we take upward as positive direction, so acceleration is negative).

The initial velocity along the vertical direction is

$v_{y0} = v_0 sin \theta = (26 m/s)(sin 30^{\circ})=13 m/s$

Now we can consider the motion along the vertical direction only. the vertical velocity at time t is given by:

$v_y(t)=v_{y0} +at$

At the point of maximum height, $v_y(t)=0$, so we can find the time t at which the ball reaches the maximum height:

$0=v_{y0}+at\\t=-\frac{v_{y0}}{a}=-\frac{13 m/s}{-9.8 m/s^2}=1.33 s$

And now we can use the equation of motion along the y-axis to find the vertical position of the ball at t=1.33 s, which corresponds to the maximum height of the ball:

$y(t)=v_{y0}t + \frac{1}{2}at^2=(13 m/s)(1.33 s)+\frac{1}{2}(-9.8 m/s^2)(1.33 s)^2=8.6 m$