Question

In: Physics

A softball is hit over a third baseman's head with speed v0 and at an angle...

A softball is hit over a third baseman's head with speed v0 and at an angle θ from the horizontal. Immediately after the ball is hit, the third baseman turns around and runs straight back at a constant velocity V=7.00m/s, for a time t=2.00s. He then catches the ball at the same height at which it left the bat. The third baseman was initially L=18.0m from the location where the ball was hit at home plate.  

1) Find a vector expression for the velocity v⃗  of the softball 0.100 s before the ball is caught.

Use the notation vx, vy, an ordered pair of values separated by a commas Express your answer numerically in meters per second to three significant figures.

2) Find a vector expression for the position r⃗  of the softball 0.100 s before the ball is caught.

Use the notation x, y, an ordered pair of values separated by a comma, where x and yare expressed numerically in meters, as measured from the point where the softball initially left the bat. Express your answer to three significant figures.

Solutions

Expert Solution

the initial velocity of the ball in y direction is

vy = g/t

=9.8/1.0 s

= 9.8 m/s

here time take 1.0 s since  it takes 2 seconds for the ball to go up and then back down,

For the x velocity, the third baseman runs for 2 seconds at a speed of 7 m/s.

d= vt = 7 m/s * 2 s = 14 m

total distance = d+ L = 14 m + 18 m = 32 m

The ball covered this distance in 2 seconds, so it must have been going 16 m/s in the x direction.

v= sqrt vx^2 + vy^2 = sqrt ( 16^2 + 7^2 ) = 18.77 m/s

The only component of velocity will change from start to finish is the y velocity. Since it takes 1 second for the ball to reach its maximum height .

it also takes 1 second for it to fall back down. This means that 0.1 s before the ball is caught, it will have been falling for 0.9 s, and

so it’s speed in the y direction is

vy = (0.9 * 9.807) = -8.8263 m/s.

vx= 16 m/s , vy = -8.82m/s

(2)

Since the ball will have been moving for 1.9 s

the position covered in x direction is

xi = 16 m/s ( 1.9)= 30.4 m

Apply kinematic equation

Vf^2 = Vi^2 + 2a(Δs)
9.807^2 = 8.8263^2 + 2 * 9.807 * Δs
96.2 = 77.9 + 19.614 * Δs

Δs = 18.3 / 19.614
Δs = 0.933 m

(x,y) = (30.4, 0.932 m)


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