SOMEONE PLEASE HELP IN PHYSICS!!!?

1) A) The pitch is horizontal ... but it doesn't say to ignore the affect of gravity ... so assume the ball travels horizontally to the bat so that the momentum vector of the ball to the bat remains horizontal

Now consider the vertical motion of the ball to find the speed the ball came off the bat

s = 37.0 m ... [taking UP as the positive direction]

a = -g = -9.8 m/s²

v(i)(vert) = ?

v(f)(vert) = 0 m/s ... [b/c at max height the ball is at rest for an instant]

v(f)² = v(i)² + 2as

v(i) = √[v(f)² - 2as]

v(i) = √[0 - 2 * (-9.8) * 37.0]

v(i) = 26.9 m/s



Now I = Δp = p(f) - p(i) = m * v(f) - m * v(i)

Draw the vector triangle for the change in momentum at the bat:

Vertical vector for p(f) with magnitude 0.160 * 26.9 = 4.30 Ns

Then from the head of that vector draw the horizontal vector pointing LEFT for -p(i) (b/c we want to add the negative of the initial momentum to the final momentum ... same as subtracting it) with magnitude 0.160 x 34.5 = 5.52 Ns


Δp is the hypotenuse of the momentum vector triangle

so Impulse = Δp = √[4.30² + 5.52²] = 7.00 Ns


Also Impulse = F Δt

Δt = 2.5 ms = 2.5 x 10^(-3) s

so 7.00 = F * 2.5 x 10^(-3)

F = 2800 N

so the magnitude of the average force the bat exerts on the ball is 2800 N



B) Impulse has the same direction as the average force ... so the average force on the ball is in the same direction as Δp

mark the angle between the vertical vector and the hypotenuse as θ

θ = arctan (5.52 / 4.30)

θ = 52°

so the direction of the average force of the bat on the ball is at an angle of 52° from the vertical TOWARDS the pitcher







2) Your answer to the first part has the right magnitude and correct number of sig figs ... and provided you took the positive direction as UP the sign of your answer is also correct

A) Just b4 hitting the ground you would have found the velocity of the jumper to be -7.4 m/s [taking UP as the positive direction]

The jumper comes to rest and in coming to rest his knees bend 40 cm = 0.4 m

so his body travels thru a distance s = 0.4 m

In that distance he slows from -7.4 m/s to 0 m/s ... so the average velocity during the stopping distance = (-7.4 + 0) / 2 = -3.7 m/s

time taken = distance traveled / speed

so Δt = 0.4 / 3.7 = 0.1 s

Impulse = F Δt

so 470 = F * 0.1

F = 4700 N


NOW ... that is the resultant force acting up and they want the force exerted by the ground on the person's feet

so F = force exerted by ground on feet - weight of person

F = F(ground on feet) - mg

F(ground on feet) = F + mg

F(ground on feet) = 4700 + 64 x 9.8

F(ground on feet) = 5327 N = 5000 N (to 1 sig fig)

so the average force exerted by the ground on the person's feet is 5000 N vertically up...Show more