Problem 2 [25 pts] A ball is thrown at a vertical brick wall. The ball is initia

Question

Problem 2 [25 pts] A ball is thrown at a vertical brick wall. The ball is initially 1.5 m above the horizontal ground, 10 m from the wall and is thrown toward the wall at an angle of 40o with respect to the ground. It takes 1.0 second for the ball to reach the wall. (a)[7 pts] What is the initial speed of the ball? (b)[7 pts] At what distance above the ground does the ball hit the wall? (c) [7 pts] What is the ball’s velocity (magnitude and direction) when the ball hits the wall? (d) [4 pts] Did the ball reach its maximum height before it hit the wall?

Explanation / Answer

Initial speed of ball = vo = ?

Direction = 40 degree with respect to horizontal

Initial horizontal speed =vox = vo cos 40

Initial vertical speed = voy = vo sin 40

(a)Time taken by ball to cover 10 m horizontal distance = t = 1.0 s

Horizontal distance = Horizontal speed x time taken

10 = vox (1.0)

10 = vo cos40

vo = 13.05 m/s

(b) Ball hits wall in 1.0 s

In 1.0 s vertical distance covered by ball = y

y = voy(t) + 0.5 at^2

y = 13.05 sin 40 (1.0) +0.5(-9.81)(1.0)^2

y = 3.48 m

Ball hits the wall 1.5 m +y m above ground i.e 1.5 m + 3.48 m = 4.98 m

(c) After 1.0s, ball has same horizontal velocity vx = vox = vocos40 = 13.05 cos 40 = 9.997 m/s

Vertical velocity of ball after 1.0 s = vy = voy +at = 13.05 sin 40 + (-9.81) (1.0) = -1.42 m/s

Magnitude of net velocity while hitting wall is v = ( vx^2 +vy^2)^1/2 =10.1 m/s

Direction = arctan(vy/vx) = arctan(-1.42/9.997) = -8 degrees i.e. 8 degrees below horizontal

(d) Since ball has vertical velocity in downward direction while hitting ball, this means it is in return journey. It has already achieved its maximum height

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