Question

Oscar throws a ball towards a wall with a speed of 25m/s at an angle of 40degrees above ground. The wall is 23m from the release of the balls
A.How far form the release point does the ball hit the wall?
B. What is the magnitude and Direction of the balls velocity as it hits the wall?
C. When it hits the wall, has it passed the highest point of its trajectory?

Answer 1

A.

$$t=\frac{d}{v\cos{40\degree}}=\frac{23}{25\cos{40\degree}}=1.2\ s.$$

$$h=v\sin{40\degree}t-0.5gt^2$$

$$h=25\sin{40\degree}(1.2)-0.5(9.8)(1.2)^2=12.2\ m$$

The distance:

$$s=\sqrt{12.2^2+23^2}=26\ m$$

B.

$$V_x=25\cos{40\degree}=19.15\frac{m}{s}$$

$$V_y=25\sin{40\degree}-(9.8)(1.2)=4.31\frac{m}{s}$$

Magnitude:

$$V=\sqrt{19.15^2+4.31^2}=20\frac{m}{s}$$

Direction:

$$\theta=\arctan{\frac{4.31}{19.15}}=13\degree$$

C. No. It is because after passing the highest point of its trajectory its velocity should be pointed below the horizont:

$$\theta<0$$