Assuming that we are talking about Earth's atmosphere (and which a = -9.8) and s = 0, this becomes:

s(t) = -4.9t^2 + 17t.

From here, part a) can be solved by finding the y-coordinate of the
vertex, part b) can be solved by finding the x-coordinate of the vertex,
the answer to part c) is the same to that as part b) as it takes the
same amount of time for it to fall back down, and part d) can be solved
by differentiating and finding the slope of the tangent line at the
point the ball hits the ground.

I hope this helps!

May 25th, 2015

(a) V² = U² +2*a*s
0 = 17² + 2*(-9.81) *s
0 = 289- 19.62*s
19.62s = 289
s = 289/19.62
s = 14.73 m
Height achieved = 14.73 m

b) V = U + a*t
0 = 17 + (-9.81) t
9.81t = 17
t = 1.73 seconds.

c) Time up = time down
Answer = 1.73 seconds

d) It will hit the ground with the same velocity with which it was thrown = 19 m/s

May 25th, 2015

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