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A ball bearing is fired straight up at a speed of 40m/s. Calculate the time during which it is in the air. Ignore air resistance.

I have been given the following equations:
[tex]v = u + at\\s = 1/2(u + v)t\\s = ut + 1/2at^2\\s = vt-1/2at^2\\v^2 = u^2 + 2as[/tex]

An explanation would be very much appreciated! Thank you.

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Answer:

t = 8.2 sec

Step-by-step explanation:

the ball will move with uniform deceleration due to gravity until it reaches the highest point at which v = 0 then will change its direction and continue with uniform acceleration until it hits the ground.

the time the ball takes to reach the highest point equals the time it takes to return to the ground again.

to calculate the total time, we calculate the first half then multiply it by 2

we use the first equation because we have all the variables except time

v = 0 , u = 40 , a = -9.8

0 = 40 - 9.8t

t = 4.1 sec

total time = 4.1 x 2 = 8.2 sec

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