Use the model for projectile motion assuming there is no air

Use the model for projectile motion, assuming there is no air resistance. A baseball player at second base throws a ball 90 feet to the player at first base. The ball is released at a point 5 feet above the ground with an initial velocity of 60 miles per hour and at an angle of 11 degree above the horizontal. At what height does the player at first base catch the ball? (Round your answer to three decimal places.) feet

Solution

We will solve the above situation for motion along the x direction and that along y direction independently.

Apparently, the ball will suffer no acceleration while travelling horizontally. However, it will have an acceleration downwards, equal to g, while travelling in the vertical direction.

Now, let us consider that the baseball is caught by the player at first base after time t seconds.

We have Vx = 60 cos11 Miles/hr = 88 cos11 = 86.3832 Feet/sec

Vy = 88 sin11 = 16.7912 feet/sec

So, Vx t = Distance = 90

or t = 90 / Vx = 90 / 86.3832 = 1.04187 seconds

We will use the this time for the motion along the vertical direction to determine the displacement along the vertical.

S = Vyt - 0.5*gt^2 = 16.7912 x 1.04187 - 0.5 x 32.2 x 1.04187^2

S = 17.4943 - 17.4764 Feet = 0.0179 feet above the launch height

Therfore the height at which the player at first base catches the ball is 5.0179 feet above the ground.