A major leaguer hits a baseball so that it leaves the bat at a speed of 32.0 m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 32.0 m/s and at an angle of 35.7 above the horizontal. You can ignore air resistance.

A)At what two times is the baseball at a height of 9.00 m above the point at which it left the bat? t1_____ t2____

B)Calculate the horizontal component of the baseball's velocity at each of the two times you found in part A. Vh1_____Vh2_____

C)Calculate the vertical component of the baseball's velocity at each of the two times you found in part A. Vv1____Vv2

D)What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat? V=

E)What is the direction of the baseball's velocity when it returns to the level at which it left the bat?

Explanation / Answer

so the ball is hit at an initial velocity, Vo=32 m/s at an angle =35.7 above the horizontal. therefore there will be two components of V:

vertical component: Vy = V1 = V sin = 32m/s sin 35.7 = 19m/s

horizontal component: Vx = V2 = V cos = 32m/s cos 35.7 = 25.7m/s

something to remember here is that the horizontal component will always be constant = 25.7m/s because there isn't any acceleration or deceleration in horizontal movement. the vertical component will take the ball to the height of 9m vertically.

.: initial vertical component of velocity=19m/s, height h=9m, a= -g = -9.8m/s2

using the equation h=vt+1/2gt2

9=19t-4.9t2

4.9t2 + 9 - 19t = 0, solve using the quadratic formula yields two times.

t = .55s and t = 3.33s

the ball will reach a height of 9m in .55s during it's ascension and 9m in 3.33s during its descent. it'll reach a point where the vertical component will be = 0. so let's name that point V3 at the top of the projectile.

total height covered by the ball:  and the time taken to reach this peak, if you want to call it that:  from the level at which the bat hit it

so the time it takes to reach the peak from 9m from the level at which the bat hit the ball

=1.94s - .55s=1.39s

Once the ball reaches the peak it has constant horizontal speed of 25.7m/s. but gravity is going to pull the ball down and this is why the ball will descend the same parabolic path it did during its ascension.

at the peak the balls initial downward speed u=v3=0. to reach a height 9m above the level it was hit by thee bat, the ball will fall: h-9 = 18.23m - 9m = 9.23m (from the peak). so time (T) it takes for the ball to fall 9.23m from the peak is: 9.23 = V3t +1/2gT2

9.23 = 0 + 4.9T2 : T=1.37s

So the total time taken to reach height 9m above the level at which it was hit by the bat during descending=time taken to ascend up to 10m + time taken to reach peak from 9m height + time taken to descend to height 9m during the projectile motion: .55+1.37+1.37=3.29s

so your answers are:

a) t=.55s and 3.33s at height 9m above the level at which it was hit by the bat.

b) your horizontal components of the velocity = 25.7m/s in both the cases.

vertical component: ascending V1 - gt = 19-(9.8)(.55)=13.61m/s

vertical component: descending V1 - gt = 1.39s x 9.8 =13.61m/s

therefore your vertical components will be 13.61m/s and horizontal 25.7m/s at the points (ascending, descending)

c) the magnitude of the velocity at the level it was hit is the same as it was leaving the ball = 32 m/s and the d) direction is 36.5 degrees clockwise.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.