A baseball is struck by a bat 46 cm away from the axis of rotation when the angular velocity of the bat is 70 rad/s.
If the ball is hit at a height of 1.2 m at a 45° angle, will the ball clear a 1.2 m fence 110 m away?
(Assume that the initial linear velocity of the ball is the same as the linear velocity of the bat at the point at which it is struck)
Help please!
1 Answer
-
To start, the linear velocity of the ball at time t=0 would be:
a) v = r*w =.46m * 70 rad/s = 32.2 m/s.
Next, since you know the ball is going upwards at 45°, then the vertical velocity is:
vy = 32.2 * sin 45° = 22.7688 m/s
Gravity applies a constant acceleration of -9.81 m/s² on vy, so you have:
b) 1/2gt² + vt + y = 0
1/2(-9.81)t² + 22.7688t + 1.2 = 0
I don’t have a calculator on me, and I don’t feel like using the quad formula in the Windows one, so I’ll just call the number you get for t n.
Next, since you have time n and the horizontal velocity (vx = vy), you just need:
c) x = vx * n
From here you should probably be able to do the last part. Good luck.
-
You don’t need the quadratic formula;
After obtaining the velocity [answer to a).] like in the previous answer (v = rw = .46m * 70rad/s = 32.2m/s) just use these two simple formulas to get the total time and range [answers b). and c).]:
b). t = 2(v)(sin(theta))/g = 2*32.2*sin(45)/9.81 = 4.6s
c). R = v^2 (sin 2*theta)/g = (32.2m/s)^2 (sin 2*45)/9.81m/s^2 = 105.7m
d). The projectile will only travel 105.7m, short of 110m.
(g = acceleration due to gravity; t = total flight time; theta = given angle of 45 degrees)
