2. A ball is thrown from a point 1.00 m above the ground with an initial velocit
ID: 1778543 • Letter: 2
Question
2. A ball is thrown from a point 1.00 m above the ground with an initial velocity 19.6 mls at an angle of 30.00 above the horizontal. a. What is the maximum height (above the ground) that the ball reaches? b. What is the velocity (magnitude and direction) of the ball when it is at this maximum height? c How much time elapses from when the ball is thrown to when it returns to 1.00 m above the ground? What is the velocity (magnitude and direction) of the ball when it returns to 1.00 m above the ground?Explanation / Answer
initial velocity vo = 19.6 m/s
along vertical
initial velocity voy = vo*sintheta = 19.6*sin30 = 9.8 m/s
acceleration ay = -g = -9.8 m/s^2
initial position y0 = 1 m
at the maximum height final velocity vy = 0
final position y = h
displacement = y - y0 = h - 1
vy^2 - voy^2 = 2*ay*(y-y0)
0 - 9.8^2 = -2*9.8*(h-1)
maximum height = h = 5.9 m <<<---ANSWER
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(b)
at the maximum height
along vertical vy = 0
along horizontal
acceleration ax = o
vx = vox + ax*t
vx = vox = vo*costheta = 19.6*cos30 = 16.9 m/s
at the maximum height velocity v = sqrt(vx^2+vy^2) = 16.9 m/s
direction = along horizontal
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c)
along vertical
initial position y0 = 1 m
final position y = 1 m
initial velcoity voy = 9.8 m/s^2
y - y0 = v0y*t + (1/2)*ay*t^2
0 = 9.8*t - (1/2)*9.8*t^2
t = 2 s
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(d)
along horizontal
acceleration ax = 0
velocity remains same at all times
vx = vox = 16.9 m/s
along vertical
vy = voy + ay*t
vy = 9.8 - (9.8*2) = -9.8 m/s
magnitude = sqrt(vx^2+vy^2) = 19.6 m/s
direction tan^-1(vy/vx) = 30 degrees below the horizontal
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