Question

Interpreting Data: A baseball is hit straight upat

an ititial velocity of 30 m/s. If the ball has a
negativeacceleration of about 10 m/s 2, how long does the ball take
toreach th top of it's path?

Answer 1

3 second.

Step-by-step explanation:

In the given question, we have given u = initial velocity = 30 m/s, acceleration = -10 m/s 2, final velocity will be zero as the balls stops after sometime. We have to find the time,

Time can be calculated by the following formula:

v = u + at

t =
`(v-u)/(a)`

t =
`(0-30)/(-10)`

t = 3 sec.

Thus, the answer is 3 second.

Answer 2

The baseball will 3 second to reach the top of it's path

Step-by-step explanation:

Initial velocity of the ball= u = 30 m/s

Ball is accelerating against the gravity =
`a = -10 m/s^2`

final velocity of the ball when it reaches to the top path= v = 0

Using first equation of motion:

v = u +at

`t=(v-u)/(a)`

`t=(0-30 m/s)/(-10 m/s^2)=3 s`

The baseball will 3 second to reach the top of it's path