distance and max height of a baseball

Question

A ballplayer standing at home plate hits a baseball that is caught by another player at the same height above the ground from which it was hit. The ball is hit with an initial velocity of 18.0 m/s at an angle of 60.0° above the horizontal.

How high will the ball rise than where it was hit?

At what distance from the home plate will the fielder be when he catches the ball?

Answer 1

Since it is launched and caught at the same elevation (height)

this is a LHR (Level Horizontal Range) problem.

There are a few different ways to solve this.

I do not know what you are studying at this time to say which is the best for you.

One way is:

This is the longest but most basic way and uses only basic principles.

1) Since it takes the same time to go up as to come down

you can break the problem into 2 halves.

-- the Up problem and the Down problem

a) The time up = time down.

b) The velocity at the top is 0.

c) The velocity when caught is the same as that when launched (hit).

d) There is a force (gravity) in the vertical direction

so there is acceleration in the vertical (up/down) part of the problem.

e) The vertical acceleration everywhere (even at the top) is a_g = g = -9.81 m/s²

f) There is no force so no acceleration in the horizontal direction after it is hit

so the horizontal velocity is constant.

g) ** As with any falling object or free fall problem

If you can find the time, you can find everything else.

2) Decompose the initial velocity into the horizontal and vertical components.

3) Use the initial vertical velocity to solve the time up or the time down.

Find either one, which every is easier for you.

Solve using the rising or falling problem - use the vertical velocity component.

Use the definition of acceleration: a(avg) = Δv/Δt where a = g

4) The total time in the air is twice the up or down time.

5) Find the distance it goes using the definition definition of velocity.

Use the total time and use the horizontal component of the velocity.

6) optional - if asked

Find the height using the definition of velocity: v(avg) = Δx/Δt

where height = Δx and use the vertical velocity component.

Since the acceleration is constant v(avg) is just the average: v(avg) = 1/2(vi+vf).

Another way is to use the equations for a LHR problem:

ONLY IF you have been taught them!

1) T(total) = 2• [ -v(i) sin(θ) / a(g) ]

2) Range = v(i)² sin(2θ) / a(g)

3) Height = | [v(i)sin(θ)]² / 2•a(g) |