Physics Projectile Motion

Question

A ball player hits a home run, and the baseball clears a wall 8.00 m high located 149.2 m from home plate. The ball is hit at an angle of 35.1 ◦ to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 1.2 m above the ground. What is the initial speed of the ball? The acceleration due to gravity is 9.81 m/s 2 . Answer in units of m/s.

Answer 1

We will use the first kinematic equation for this one.

x = x0 + vt + 1/2 at^2

We must split this into a vertical component and a horizontal component.

Let's start with the horizontal.

x = 149.2 (distance travelled)

x0 = 0

v = v(cos(35.1)) This is the horizontal component of velocity. Draw a right triangle to see why.

a= 0

Our goal is to determine v in terms of t.

149.2 = 0 + v(cos(35.1)) t

149.2 = v*0.818*t

v = 182.4/t

Next we do the same for the y direction.

x = 8 (height of the wall)

x0 = 1.2 (starting height)

v = v(sin(35.1))

a - -9.81 (don't forget the negative)

8 = 1.2 + v(sin(35.1))t + 1/2 (-9.81)t^2

6.8 = 182.4/t * 0.575 * t - 4.9t^2

6.8 = 104.9 - 4.9t^2

t = 4.474

v = 182.4/t

v = 182.4/4.474

v = 40.77 m/s