Projectile Passing Two Walls
A ball is projected from point O on horizontal ground with speed 20 m/s at 35 degrees above the horizontal. A vertical wall of height 5 m stands 18 m from O. A student calculates the time to reach the wall from horizontal motion, then substitutes into vertical motion, obtaining a height of about 6.7 m, and concludes that the ball clears the wall. The teacher asks for a fuller analysis: the speed and direction of the ball when it returns to the ground, and whether it would clear a second wall of height 6 m placed 30 m from O. Use g = 9.8 m/s^2 and ignore air resistance. Give answers to 3 significant figures. The student must keep horizontal and vertical motion separate and must check that the ball is still in flight at the second wall.
1. Show that the ball clears the first wall. Include the time and vertical height at the wall.
Read the worked answer
Resolve the initial velocity: u_x=20cos35=16.383 m/s and u_y=20sin35=11.472 m/s.
Time to the wall: t=18/16.383=1.099 s. Vertical displacement: y=u_y t - (1/2)g t^2 = 11.472(1.099)-4.9(1.099)^2 = 12.61-5.92 = 6.69 m.
Since 6.69>5, the ball clears the 5 m wall. The horizontal speed is constant, while the vertical motion has acceleration -9.8 m/s^2.
2. Find the speed and direction of the ball's velocity when it returns to ground level.
Read the worked answer
Time of flight: T=2u_y/g=2(11.472)/9.8=2.341 s.
Horizontal velocity remains v_x=16.383 m/s. Vertical velocity at ground level: v_y=u_y-gT=-11.472 m/s.
Speed = sqrt(v_x^2+v_y^2)=sqrt(16.383^2+11.472^2)=20.0 m/s. Direction: tan(theta)=11.472/16.383, so theta=35.0 degrees below the horizontal. This agrees with symmetry: the ball returns to the same height with the same speed and with the vertical component reversed.
3. Determine whether the ball would clear a second wall of height 6 m at horizontal distance 30 m. Check first that the ball is still in flight.
Read the worked answer
Range: R=u_x T=16.383(2.341)=38.36 m, so at x=30 m the ball is still in flight.
Time to x=30: t=30/16.383=1.831 s. Height: y=11.472(1.831)-4.9(1.831)^2 = 21.00-16.43 = 4.57 m.
Since 4.57<6, the ball does not clear a 6 m wall at 30 m; it would hit the wall. The check that R>30 is essential, otherwise the projectile formula at x=30 would not apply.