Question

during a baseball game, a batter hits a foul ball down the left field foul line. while being hit 0.762 m above the ground, the ball was given a velocity of 33.53 m/s at an angle of 55 degrees above the ground. 5 seconds after being hit, the ball was seen clearing a 11.28 m high wall in left field.

a) what is the horizontal distance from home plate to the left wall?

b) what is the vertical distance of the ball above the ground as it travels over the wall?

Please show me how to do this, I'm having a rough time figuring it out.

Answer 1

Let´s stablish the reference system to do the calculations:

+y direction will be vertical, upwards

+x direction will be horizontal

To answer this problem we need to visualize the movement of the ball in these two directions. In the +x direction, it is a constant velocity movement, while in the +y direction, the ball is influenced by the gravity. In other words, the horizontal (+x) component of the velocity of the ball is constant, and the vertical (+y) direction of the speed of the ball is constantly changing.

In question a), we need to determine the horizontal distance that the ball travelled in 5 seconds. Since the horizontal component of the velocity, v_x, is constant, this is a fairly simple calculation, given by

, where "x" is the horizontal distance travelled by the ball, t is the time and v_x is the horizontal component of the velocity, which can be determined by

,

where "v" is the total magnitude of the initial speed, in this case, v= 33.53 m/s

And "" is the angle between the velocity vector and the horizontal direction, in this case = 55º.

Then,

And the horizontal distance from the home plate to the wall is

(ANSWER TO QUESTION a)

For question b) we need to determine the height at which the ball is when it has been in the air during 5 seconds.

We know that the vertical component of the velocity, v_y is constantly changing due to the gravity, so we can use the following expression:

where "y₀" is the initial height of the ball at the start of the movement, in this case y₀= 0.762 m

"v_y0" is the initial velocity of the ball in the +y direction, in this case

"a" is the acceleration affecting the component of the velocity involved in the calculation, in this case, a is the acceleration of gravity, and since the +y direction points upwards, a= -9.81 m/s

Then, the vertical distance of the ball above the ground as it travels over the wall is

ANSWER TO QUESTION b)