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?

Respuesta :

Answer:

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

Explanation:

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

Ball is accelerating against the gravity = [tex]a = -10 m/s^2[/tex]

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

Using first equation of motion:

v = u +at

[tex]t=\frac{v-u}{a}[/tex]

[tex]t=\frac{0-30 m/s}{-10 m/s^2}=3 s[/tex]

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

Answer:

3 second.

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 = [tex]\frac{v-u}{a}[/tex]

t = [tex]\frac{0-30}{-10}[/tex]

t = 3 sec.

Thus, the answer is 3 second.