1-What is the instantaneous velocity of a freely falling object 6 s after it is

Question

1-What is the instantaneous velocity of a freely falling object 6 s after it is released from a position of rest?
2-A car takes 14 s to go from v = 0 to v = 70 m/s at approximately constant acceleration. If you wish to find the distance traveled using the equation d = 1/2 at2, what value should you use for a?

3-Surprisingly, very few athletes can jump more than 2.2 ft (0.66 m) straight up. Use d = 1/2 gt2 and solve for the time one spends moving upward in a 2.2 foot vertical jump. Then double it for the "hang-time" -- the time one's feet are off the ground.

Explanation / Answer

1 the object starts with a velocity of 0. Every second it is going 9.8 m/s faster than the second before. Therefore, after 6 seconds, it is going 58.8 m/s. 2 the change in velocity is 70 m/s. The change in time is 14s. So the average acceleration is (70m/s)/14s = 5 m/s^2 3 d = 1/2 gt2 t2 = 2d/g t = SQRT(2d/g) = SQRT(2*0.66m/9.8m/s^2) = 0.367 s Hang time is 0.734s

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