Question

8. A tennis ball is thrown straight up with an initial speed of 22.5 m/s. It is
caught at the same distance above the ground.
a. How high does the ball rise?
b. How long does the ball remain in the air?
​

Answer 1

Answer:

1) 25.8m

2) 4.6

Step-by-step explanation:

1) (0m/s)power 2- (22.5m/s)power 2= 2(-9.8m/s power 2) y

(-506mpower/spower2) =

(-19.6m/s power 2)y

y=25.8m

2)0wholem/a = 22.5wholem/s +( -9.8 whole m/a power 2)( t)

-22.5 m/a = -9.8 m/spower2 (t)

t= 2.305

(2.35) (2)= 4.6

Answer 2

Given: A tennis ball is thrown straight up with an initial speed of 22.5 m/s. It is caught at the same distance above the ground.

To find: 1- How high does the ball rise?

2- How long does the ball remain in air

Solution: Let the initial velocity of the ball be u, final velocity be v, distance travelled be s and time taken to go upwards be t.

Now, u = 22.5 m/s

v = 0 (at highest point, velocity of ball becomes 0)

g(during upward motion) = -10 m/s^2

Using kinematics formula:

${v}^{2} - {u}^{2} = 2gs$

$= > 0 - ( {22.5)}^{2} = 2 \times - 10 \times s$

=> - 506.25 = -20 s

=> s = 506.25/20

=> s = 25.3125 m

Now, while going upward, time taken can be given by formula:

v = u +gt

=> 0 = 22.5 -10 t

=> t = 22.5/10

=> t = 2.25 seconds

Time taken for a body to go up to the highest point is equal to the time taken to fall down from that point.

Total time the ball remains in air

= Time taken in going up + Time taken in coming down

= 2.25 + 2.25

= 4.5 seconds

Therefore, the ball rises to a height of 25.3125 m and the ball remains in air for 4.5 seconds.