well, the height at time t is
h(t) = 0 + 60 sin42° - 4.9t^2
assuming it is hit from a height of 0 (unusual for a baseball...)
anyway, if needed plug in the initial height and then just work with the quadratic function the way you remember from Algebra I
(recall that the vertex is at x = -b/2a)
A batter hits a baseball so that it leaves the bat at a speed V0 = 60.0 m/s at an angle of 42 degrees as measured from the horizontal. Take g to be 9.8 m/s^2 as usual.
a. Find the position of the ball, and the magnitude and direction of its velocity at t=2.0 sec.
b. Find the time when the ball reaches the highest point of its flight
c. Find its height h at this point and the acceleration at this point.
d. Find the horizontal range, R, the horizontal distance from the starting point to where the ball hits the ground.
1 answer
0 answers