Question

An ball is thrown upward with a velocity of 60 m/sec. The acceleration is v′(t) = −9.8 m/sec2. Assume that the thrown ball is at height 0 metres at time t = 0.
(a) when does it reach highest point?
(b) whe does it hit the ground?
(c) how high is its highest point?
(d) how fast is it travelling the instant before it hits the ground?

Answer 1

My answers using 'applications of integration'
A) v= -9.8 t + C
Therefore C=60 and t=6.12s
Thus it takes 6.12s to reach its highest point

B) the above times 2 = 12.24s

C) antiderivative of velocity for height/distance
Therefore =-4.9t^2 + 60t + c
Sub in 0 we get c=0

Max height sub time into equation
=183.67m

D) I am stuck on this one but have used formula
V^2=u^2+2as
V=84.87m/s


Are my answers correct and if not can someone please guide me?

Answer 2

man it's difficult to do math when you're hungry :DDD

Answer 3

Yes it looks good except d. The answer should be 60 m/s.

Answer 4

\[v _{f}= -9.8t + 60 \]

Answer 5

Just plugin your answer for b.

Answer 6

The equation you used for d works however the initial velocity would be zero so it just reduces to
\[v _{f}= \sqrt{2g(h)}\]

Answer 7

I hope this helps :)

Answer 8

Why would it be 60? Does the ball not fall faster on the way down or would it be the same as the way up as per the question with acceleration being 9.8m/s for the way down?

Answer 9

My calc for d is
V^2=60^2+2(9.8)(183.67)
=84.87m/s

Answer 10

At the top of the parabola you have zero velocity as you are throwing it straight up. So, it is the same as dropping the ball from that height and then falling under the force of gravity.

Answer 11

I phrased it a little wrong up top its to late :P. You do have an initial velocity but if you look at the way I described above it is the same.

Answer 12

Yep I get it, thank you