Question #141952
A ball falling freely in the field of gravity passed the final section of its trajectory of length 45 m during time interval 1 sec. Find the
height from which the ball fell if its initial velocity was zero. Give your answer in meters m]. Assume the free fall acceleration to be
10 m/s2. The air resistance is negligible.
1
Expert's answer
2020-11-03T10:28:29-0500

The distance trevelled by the free falling ball is:


dlast=v0t+gtlast22d_{last} = v_0t + \dfrac{gt_{last}^2}{2}

where v0v_0 is the initial speed of ball at the beginning of the last section, g=10m/s2g = 10m/s^2 is the free fall acceleration, tlast=1st_{last} = 1s is the time of covering the last section, dlast=45md_{last}= 45m is the length of the last section.

Expressing v0v_0, obtain:


v0=dlastgtlast22=4510122=40m/sv_0 = d_{last} - \dfrac{gt_{last}^2}{2} = 45-\dfrac{10\cdot 1^2}{2} = 40m/s

In order to gain such velocity from the rest, the ball should have spent the following amount of time:


t=v0g=4010=4st = \dfrac{v_0}{g} = \dfrac{40}{10} = 4s

In this time it covered the following distance:


d=gt22=10422=80md = \dfrac{gt^2}{2} = \dfrac{10\cdot 4^2}{2} = 80m

Thus, the total height from which the ball fell is:


h=d+dlast=80m+45m=125mh = d + d_{last} = 80m + 45m = 125m

Answer. 125 m.


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