You throw a ball toward a wall at speed 20 m/s and at angle theta=60 degrees abo

Question

You throw a ball toward a wall at speed 20 m/s and at angle theta=60 degrees above the horizontal. The wall is a distance d= 30 m from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall? (d) What is the angle when it hits the wall? (e) When it hits, has it passed the highest point on its trajectory? (f) If it bounces back from the wall with the same speed and angle, how far will the ball hit the ground again?

Explanation / Answer

The ball is launched with a horizontal velocity of 20*Cos60 = 10 m/s. At this speed, it will cover the horizontal distance of 30 m in 3.0 s. It starts with a vertical velocity of 20*Sin60 = 17.32m/s. Using v = u + at, with a = -g, we can work out that it takes 1.6 s to reach maximum height. Thus it hits the wall on its way up. We need to work out its vertical height, while travelling upwards after 1.15 s. h = ut + 1/2*a*t^2 h = 17.32*3.0 + 0.5*(-9.8)*(3.0^2) h = 7.86 m e) no

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