Question

A ball is thrown upwards at 8m/s. If the point of release is 2.1 m above ground, how long does it take for the ball to hit the ground?

Answer 1

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Answer 2

Suppose the ball is thrown upwards from the point A. It reaches the highest point B (where its velocity becomes 0) and then again returns to point A. From here it continues to travel and finally reaches point C at ground.

\(\sf Step~1\)
Let us calculate the time it will take to reach point B from A
Given,
Initial Velocity(u)=8m/s
Final velocity should be equal to 0m/s
acceleration in this case will be -9.8m/s

Now, using v=u+at..
0=8-9.8t
=>t=0.81 seconds.
It will take same time to return back to point Afrom B. Also on its return at A its velocity will be same 8m/s (By conservation of mechanical energy).

Now using s=ut+1/2at^2
2.1=8t+0.5*9.8*t^2
=>2.1=8t+4.9t^2
=>4.9t^2+8t-2.1=0

=>t=0.23 seconds

Now total time taken=2*0.81 + 0.23 =1.85 seconds

Answer 3

Got it @steph_q ?

Answer 4

Thanks I got it! Your answer is detailed enough for me, but for others maybe you could include why you multiplied 0.81 by 2, or what a, t, and s are. When I was a beginner abbreviations that weren't explained frustrated me, so maybe for beginner's sake, but thank you very much!

Answer 5

Thanx for ur suggestion...i'll keep that in mind :)

Answer 6

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Answer 7

I have a question, why is acceleration negative 9.8 and not positive? See this example http://www.enotes.com/homework-help/ball-thrown-vertically-upward-with-speed-26-3m-s-103831 as the ball is going up and gravity is pulling it down, they used positive 9.8 as acceleration.

Answer 8

Acceleration is towards the earth and the velocity is upwrds. This is why i took acceleration -9.8. it's upon u, if u take acceleration as 9.8 then u'll have to take the velocity negative.