Answer to Question #85304 in Mechanics | Relativity for Edward

Question

A ball is thrown on level ground with a speed of 20.0 m/s at an angle of 40.0 deg with the ground. There is a 3.0 m high femce at a distance of 35.0 m from the point of release (a) will the ball hit the fence or go over it? (B) find the velocity (magnitude and direction) of the ball when it is at 35.0 m from the point of release where it hits the fence and goes over it.

Answer 1

The equation of trajectory of a ball

y(x)=y_0+x tan⁡〖θ-(gx^2)/(2v_0^2 cos^2⁡θ )〗

Therefore

y(35.0)=0+35.0 tan⁡〖40.0°-(9.81×〖35.0〗^2)/(2×〖20.0〗^2 cos^2⁡〖40.0°〗 )〗=3.77 m

Since y(35.0)>h=3.0 m, so the ball will go over fence.

The components of the ball’s velocity

v_x=v_0 cos⁡θ=20.0 cos⁡〖40.0°=15.3 m/s〗

v_y=v_0 sin⁡θ-gt

Since

x=v_x t

We obtain

t=x/v_x =35.0/15.3=2.29 s

The y-component of the ball’s velocity

v_y=v_0 sin⁡θ-gt=20.0 sin⁡〖40.0°-9.81×2.29=-9.61 m/s〗

The magnitude of velocity

v=√(v_x^2+v_y^2 )=√(〖15.3〗^2+(-9.61)^2 )=18.1 m/s

The direction

θ=tan^(-1)⁡〖v_y/v_x =tan^(-1)⁡〖(-9.61)/18.1=-28.0° or 28.0° below the horizon〗 〗

Answers:

Ball will go over fence

18.1 m/s (28.0° below the horizon)