Projectile Motion: Problem 4.32, page 85 You throw a ball toward a wall at speed
ID: 2078682 • Letter: P
Question
Projectile Motion: Problem 4.32, page 85 You throw a ball toward a wall at speed 25.0 m/s and at angle 40.0° above the horizontal (see the figure here). The wall 25.0 m/s and at angle 6o 40.0 above the horizontal (see the figure here.). The wall is distance d = 22.0 m from the release point of the ball. 1) How far above the release point does the ball hit the wall? Go 2) What are the horizontal and vertical components of its velocity as it hits the wall? 3) When it hits, has it passed the highest point on its trajectory?Explanation / Answer
v = 25 m/s
theta = 40 deg
d = 22 m
1. horizontal velocity = vcos(theta)
initial vertical velocity = vsin(theta)
let time to reach the wall be t
then vcos(theta) = d/t
t = d/vcos(theta)
so, if h is the hieght of the point where the ball strikes the wall
then h = vsin(theta)*t - 0.5*g*t^2 = vsin(theta)*d/vcos(theta) - 0.5g*d^2/v^2cos^2(theta) = tan(theta)*d - 0.5g*d^2/v^2cos^2(theta)
h = tan(40)*22 - 0.5g*22^2/25^2cos^2(40) = 11.987 m
2. when the ball hits the wall
horizontal component of velocity, vcos(theta) = 25cos(40) = 19.1511 m/s
vertical component, v = vsin(theta) - gt = 25sin(40) - g*22/25cos(40) = 4.8 m/s
3. when the ball hits the wall
we have to find theta max
let the ball hit the highest point of the wall for theta max
then
equation of trajectory of a projectile gives us
y = xtan(theta) - gx^2/2v^2cos^2(theta)
y = 16 m
x = 22 m
16 = 22tan(theta) - g*22^2/2*25^2*cos^2(theta)
16 = 22tan(theta) - 7.596sec^2(theta) = 22tah(theta) - 3.798(1 + tan^2(theta)) = 22tan(theta) - 3.798 - 3.798tan^2(theta)
3.798tan^2(theta) - 22tan(theta) + 19.798 = 0
tan(theta) = 4.67, 1.11
theta = 77.91, 47.98
so theta-max = 77.91 deg
4. when the ball goes over the wall
range, R = v^2*sin(2theta)/g = 25^2sin(2theta)/g = 26.09 m, 63.36 m
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