Abatter hits a pitched ball when the center of the ball is 1.31 m above the ground. the ball leaves the bat at an angle of 45° with the ground. with that launch, the ball should have a horizontal range (returning to the launch level) of 103 m. (a) does the ball clear a 8.55-m-high fence that is 93.0 m horizontally from the launch point? (b) at the fence, what is the distance between the fence top and the ball center?

Explanation:

Given

ball is at height of 1.31 m

Horizontal Range=103 m

launch angle

u=31.77 m/s

we know equation of trajectory of Projectile is

here x=93 m

y=9.02 m

Thus at x=93 m ball is at a height of 9.02 above its initial position i.e. 10.33 m above ground

Distance between fence and height of ball=10.33-8.55=1.78 m

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6

c)as speed changes, wavelength and frequency change.

everything changes!