A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0m/s and at an angle of 38.2degrees above the horizontal. You can ignore air resistance.

At what two times is the baseball at a height of 11.5m above the point at which it left the bat?

Calculate the horizontal component of the baseball's velocity at an earlier time calculated in part (a).

Calculate the vertical component of the baseball's velocity at an earlier time calculated in part (a).

Calculate the horizontal component of the baseball's velocity at a later time calculated in part (a).

Calculate the vertical component of the baseball's velocity at a later time calculated in part (a).

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

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

Explanation / Answer

a) y = y0 + v0y t + 1/2 a t^2

11.5 = 30*sin(38.2 degrees)*t - 0.5*9.81*t^2

t=0.781 s and 3.00 s

b) horizontal velocity doesnt change

vx = 30*cos(38.2 degrees)= 23.6 m/s

c) vy = v0 + a t = 30*sin(38.2 degrees) - 9.81*0.781= 10.9 m/s

d) vx = 23.6 m/s as found earlier

e) vy = 30*sin(38.2 degrees) - 9.81*3.00 = -10.9 m/s

f) conservation of energy so same speed

30 m/s

g) now will be downwards so -38.2 degrees

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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