velocity and acceleration?

Question

a tennis ball is thrown vertically upward with an initial velocity of +8.0 m/s
A)what will the balls speed be when it reaches its starting point
B)how long will the ball take to reach its starting point

please solve and show your work, i would like to understand how this problem works as it is the only one in the section that i don't understand

Answer 1

The ball will travel upward to apogee and return to where it was released. The speed when it returns (ignoring air resistance) will be -8.0 m/s.

The time can be calculated from v(t)=8-9.8*t
or
-8=8-9.8*t solve for t
t=16/9.8
1.63 seconds
j

Answer 2

Let me explain clearly. Split the journey of the ball into two parts:

1. UPWARD

Initial velocity = 8m/s (given)
Acceleration (downward) due to gravity = -10m/s (well-known fact)

Final velocity (just before balls turns round to come down) = 0m/s

First work out how high ball will rise (s) from

v2 - u2 = 2as
8 X 8 - 0 = 2 x(-10)s

s = 3.2 m

Then

Work out time taken to reach its peak, t, from

s = ut + 1/2 at2

3.2 = 8t + (1/2 (-10) X (-10)

8t = 50/3.2 = 15.6s
t = 15.6/8 = 1.95s
DOWNWARD

s = ut + 1/2 at2

3.2 = 0 + 1/2 X 10 X t2

t2 = 3.2/5 = 0.64
t = 0.8s

TOTAL TIME (answer to b) = 0.8 + 1.95 = 2.75s

Part a) v2 - u2 = 2as

v2 = 2 X 10 X

t2 = 15.6/5 = 0.312
t = 0.56 s

Total time = 0.56 + 1.95 = 2.51s

a)

v2 - u2 = 2as

v2 = 2 X 10 X 3.2 = 64

v = 8 m/s

Hope this helps

Answer 3

The answer is increase the velocity in a non-constant way. For example, you change the velocity first slowly, then quickly, you are always changing it, but you can vary the RATE at which you change it. You know, since acceleration= change in velocity/ time