You Throw a Ball You throw a ball toward a wall with a speed of 31 m/s at an ang
ID: 1988113 • Letter: Y
Question
You Throw a Ball You throw a ball toward a wall with a speed of 31 m/s at an angle of 40.0° above the horizontal directly toward a wall (Fig. 5-33). The wall is 22.0 m from the release point of the ball.(a) How far above the release point does the ball hit the wall?
1 m
(b) What are the horizontal and vertical components of its velocity as it hits the wall?
2 m/s (horizontal)
3 m/s (vertical)
(c) When it hits, has it passed the highest point on its trajectory?
no
not enough information to decide
yes
Explanation / Answer
(a) At the beginnning,
V0x = V cos40 = 31 cos40 = 23.747 m/s V0y = Vsin40 = 31 sin 40 = 19.926 m/s
The time the ball travels is
t = 22 /V0x= 22/23.747 = 0.926 s
Height is
h = V0yt - gt2/2 = 19.926(0.926)-9.81(0.926)2/2 = 14.25 m
(b) The horizontal velocity does not change
Vhoriz = 23.747 m/s
Vvert = V0y - gt = 19.926-9.81(0.926)=10.84 m/s
(c) It has not pass the highest point.
At highest point Vy = 0, It takes the following time to the highest point
Vy = V0y - gthigh =0 thigh = V0y/g = 2.03 s which is greater than 0.926 s. Therefore, It has not reached the highest point.
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