From a height of 25 meters a ball is thrown vertically upwards at a velocity of 5 meters per second.

Question

a. What time in seconds will the ball reach its maximum height?

b. What is its maximum height?

c. When will the ball strike the ground?

d. What will be the ball’s velocity when it strikes the ground?

can someone help me out with this?

Answer 1

a.
Vi = 5
v = Vi - 9.8 t
v = zero at top
t = 5/9.81 seconds to top

b.
h = Hi + Vi t -4.9 t^2
= 25+ 5 t - 4.9 t^2

c.
0 = 25 + 5 t - 4.9 t^2
solve quadratic for t
use the positive t
the negative t is before you threw it :)

d.
again v = Vi - 9.81 t
v = 5 - 9.81*t
use t from part c
v will be negative (down) of course

Answer 2

oh, since this is calculus and not physics:

d^2h/dt^2 = -9.81
so
dh/dt = -9.81 t + constant
at t = 0, v = Vi
so
dh/dt = Vi -9.81 t

then
h = Vi t -(1/2)9.81 t^2 + constant
h = Hi when t = 0
so
h = Hi + Vi t -4.9 t^2