Question
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.
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.
Answers
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)
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)
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