Physics 111 Projectile Motion 2.0 Practice Problem A rock is thrown with an initial vertical velocity component of 30 m/s and an initial horizontal velocity component of 40 m/s. What will these velocity components be one second after the rock reaches the top of its path? Assuming the launch and landing heights are the same, how long will the rock be in
the air? Assuming the launch and landing heights are the same, how far will the rock land from where it was thrown? Solution: A. The horizontal component of the velocity remains constant, 40
m/s. The vertical component of the velocity decreases by 10 m/s every second. So at the peak, the vertical component of the velocity is zero and one second later the vertical component of Solution:
b. Assuming the launch and landing heights are the same, how long will the rock be in the air? Since the vertical component of the velocity decreases by 10 m/s every second, it will take 3 seconds for the vertical component of the velocity to slow from 30 m/s to 0 m/s, and another 3 seconds for the rock to accelerate from 0 m/s to -30 m/s. The total time in the air is 6
seconds. Solution: c. Assuming the launch and landing heights are the same, how far will the rock land from where it was thrown? Since the rock is in the air for six seconds and the rock moves horizontally 40 meters each
second, the range for the rock is 6 s x 40 m/s = 240 m. Practice Problem 2 A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between
the table's edge and the ball's landing location. Solution: Time: Use delta y = Vyi(t) + .5(a)(t^2) (0-.60) = 0(t) + .5(-9.8)(t^2) T = .17 seconds Delta x Use equation delta x= .5( vf + vi) (t)
= .5(0+24)(.17) = 2.1 meters Problem 3 A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine: time of flight
the horizontal distance peak height of the long-jumper Solution: Time: Use vf= vi + at ( because all others have delta y which we dont have) 0 = 3.71 + (-9.8)(t) T= .37 seconds Delta
x Use equation Delta x = .5 ( vf + vi)(t) Because it doesnt have acceleration and has time (xf 0)= .5 ( 0 + 11.41)(.37) = 2.11 meters Delta y/2 Max height will be mid flight or .37/2 = .185 seconds Use equation Delta y = Vi(t) + .5(-9.8)(t^2)
Delta y = 3.71 ( .185) + .5 (-9.8) (.815^2) Delta y for half way or mazx height is .52 meters Practice 4 Alice throws the ball to the +X direction with an initial velocity 10m/s. Time elapsed during the motion is 5s, calculate the height that object is thrown and Vy component of the velocity after it
hits the ground. Answer: Problem John kicks the ball and ball does projectile motion with an angle of 53 to horizontal. Its initial velocity is 10 m/s, find the maximum height it can reach, horizontal displacement and total time required for this motion.
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