Answer to Question #262837 in Physics for Mix

Question

Ellen throws a ball toward a wall at speed 30.0 m/s and at angle 45.0° above the

horizontal. The wall is distance 25.0 m from the release point of the ball.

a) (5%) How far above the release point does the ball hit the wall?

b) (5%) What is the horizontal component of its velocity as it hits the wall?

c)

(5%) What is the vertical component of its velocity as it hits the wall?

d

(5%) When it hits, has it passed the highest point on its trajectory?

Answer 1

(a) Let's first find the time that the ball takes to hit the wall:

$x=v_{0x}t=v_0tcos\theta,$ $t=\dfrac{x}{v_0cos\theta}=\dfrac{25.0\ m}{30.0\ \dfrac{m}{s}\times cos45.0^{\circ}}=1.18\ s.$

Then, we can find how far above the release point the ball hits the wall from the kinematic equation:

$y=v_{0y}t-\dfrac{1}{2}gt^2,$ $y=v_0tsin\theta-\dfrac{1}{2}gt^2,$ $y=30.0\ \dfrac{m}{s}\times1.18\ s\times sin45.0^{\circ}-\dfrac{1}{2}\times9.8\ \dfrac{m}{s^2}\times(1.18\ s)^2=18.21\ m.$

(b) We can find the horizontal component of ball's velocity as it hits the wall as follows:

$v_x=v_0cos\theta=30.0\ \dfrac{m}{s}\times cos45.0^{\circ}=21.21\ \dfrac{m}{s}.$

(c) We can find the vertical component of ball's velocity as it hits the wall as follows:

$v_y=v_0sin\theta-gt,$ $v_y=30.0\ \dfrac{m}{s}\times sin45.0^{\circ}-9.8\ \dfrac{m}{s^2}\times1.18\ s,$ $v_y=9.65\ \dfrac{m}{s}.$

(d) As we can see from the calculations, when the ball hits the wall $v_y>0$. It means that the ball has not reached the highest point on its trajectory.