You throw a ball with a speed of 25m/s at an angle of 40 degrees above the horiz

Question

You throw a ball with a speed of 25m/s at an angle of 40 degrees above the horizontal directly toward a wall as shown in figure. The wall is 22m from the release point of the ball. A) how long is the ball in the air before it hits the wall? B.) how far above the release point does the ball hit the wall? C) what are the horizontal and vertical component of its velocity as it hits the wall? D) has it passed the highest point on its trajectory when it hits?
40° 22 m 2. You throw a ball with a speed of 25 m/s at an angle of 40° above the horizontal directly toward a wall as shown in the figure. The wall is 22 m from the release point of the ball (a) How long is the ball in the air before it hits the wall? (b) How far above the release point does the ball hit the wall? (c) What are the horizontal and vertical components of its velocity as it hits the wall? (d) Has it passed the highest point on its trajectory when it hits?

Explanation / Answer

Given : Speed = 25 m/s, Angle = 45 deg, d = 22m

Components of initial velocity and acceleration:

V0x =25 cos 400 = 19.15 m/s and ax = 0

V0y = 25 sin 400 = 16.06 m/s ay = -g = -9.8 m/s

Time to strike on the wall , t = d/v0x = 22/19.15 = 1.15 sec

Now,y = y0 + vo+ 1/2 ay2

   = (16.06)(1.15) - 1/2(9.8)(1.15)2

= 12m.

Therefore the ball hits 12m above the release point.

The x and y components of the ball's velocity

Vx = V0x + axt

= 19.2 m/s

Vy = V0y + ayt

= (16.06)(1.15) - 1/2(9.8)(1.15)

= 4.83 m/s

In the above equation the component of y is positive. Had the ball been crossed the peak of trajectory before hitting the wall the y component equation would have been negative. So, the hall didnot cross the peak of trajectory.

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