# Hello, please can someone help me with this physics question?

A batter hits a baseball so that it leaves the bat with an initial speed v0= 60.0m/s at an initial angle 42 with horizontal.

Find:

A) The position of the ball and the magnitude and direction of its velocity at =2s.

B) Find the time when the ball reaches the highest point of its flight.

C) Find the height H and the acceleration at this point.

D) Find the horizontal range R.

### 2 Answers

- Anonymous6 months ago
I'm sure there are people who can.

- 6 months ago
A) we know v0 = 60.0 m/s

angle(theta) = 42 degrees

(g) gravity = 9.8 m/s^2

We want to find the position of the ball AND the magnitude and direction of its velocity at 2s.

If we neglect air drag, the ball is a projectile for which the vertical and horizontal motion can be analyzed separately.

For position, (range) R = Vo^2 / g x 2sin theta

for the height at this position, you can use y - yo = (Vo sintheta) t -1/2 gt^2 Remember to set y0 = 0 since starting at an angle from the horizontal.

Theres the start.