A batted baseball is hit with a velocity of 47.3 m/s, starting from an initial height of 8 m. Find how high the ball travels in two cases:
(a) a ball hit directly upward and
(b) a ball hit at an angle of 70° with respect to the horizontal.
Also find how long the ball stays in the air in each case.
case a
case b
a. h = (Vf^2 - Vo^2) / 2g,
h = (0 - 2237.29) / -19.6 = 114.15m.
b. Vo = 47.3m/s @ 70 deg.
Yo = ver. = 47.3sin70 = 44.45m/s.
h = (0 - (44.45)^2) / -19.6 = 100.8m.
To solve these problems, we can use the equations of motion for projectiles. The key equations we need are:
For the upward motion:
1. h = h0 + v0t + (1/2)gt^2
2. v = v0 - gt
For the horizontal motion:
1. x = x0 + v0x t
where:
- h represents the height
- h0 represents the initial height
- v represents the velocity
- v0 represents the initial velocity
- g represents the acceleration due to gravity
- t represents the time
- x represents the horizontal distance
- x0 represents the initial horizontal position
- v0x represents the initial horizontal velocity
Let's solve each case step by step:
(a) Ball hit directly upward:
1. Calculate the time it takes for the ball to reach its maximum height:
Using equation 2, we know that the final vertical velocity at the top of the ball's trajectory is 0 m/s. Therefore, we have v = 0 m/s, v0 = 47.3 m/s, and g = 9.8 m/s^2.
Using equation 2, we can solve for t:
0 = 47.3 - 9.8t
Solving for t gives: t = 4.84 seconds.
2. Calculate the maximum height:
Using equation 1, we know that at the maximum height, the final vertical velocity is 0 m/s again. Therefore, we have v = 0 m/s, v0 = 47.3 m/s, h0 = 8 m, g = 9.8 m/s^2, and t = 4.84 seconds.
Using equation 1, we can solve for h:
0 = 8 + 47.3(4.84) - (1/2)(9.8)(4.84)^2
Solving for h gives: h = 118.96 meters.
3. Calculate the time the ball stays in the air:
Since the ball reaches its highest point at t = 4.84 seconds, and we already know that the total time of flight to reach that point is double this time: 4.84 seconds x 2 = 9.68 seconds.
Therefore, the ball travels to a height of 118.96 meters when hit directly upward, and it stays in the air for 9.68 seconds.
(b) Ball hit at an angle of 70° with respect to the horizontal:
1. Calculate the horizontal velocity:
Since the ball is hit at an angle of 70° with respect to the horizontal, we can calculate the horizontal component of the initial velocity using trigonometry:
v0x = v0 * cosθ
where θ is the angle, and in this case, θ = 70°.
v0x = 47.3 * cos(70°)
v0x = 47.3 * 0.3420
v0x = 16.17 m/s (rounded to two decimal places).
2. Calculate the time of flight:
Since the time of flight is symmetrical for the upward and downward motion, we can use the horizontal distance traveled to find the time of flight.
Using equation 1, we know that at the highest point of the ball's trajectory, the final vertical velocity is 0 m/s. Therefore, we have v = 0 m/s, v0 = 47.3 m/s sin(70°), and g = 9.8 m/s^2.
Using equation 2, we can solve for t:
0 = 47.3 sin(70°) - 9.8t
Solving for t gives: t = 4.03 seconds.
3. Calculate the maximum height:
Since we know the time of flight is 2 * t = 2 * 4.03 = 8.06 seconds, we can use this time to calculate the maximum height using equation 1.
Using equation 1, we have v = 0 m/s, v0 = 47.3 m/s sin(70°), h0 = 8 m, g = 9.8 m/s^2, and t = 8.06 seconds.
Using equation 1, we can solve for h:
0 = 8 + 47.3 sin(70°)(8.06) - (1/2)(9.8)(8.06)^2
Solving for h gives: h = 63.90 meters.
Therefore, the ball travels to a height of 63.90 meters when hit at an angle of 70°, and it stays in the air for 8.06 seconds.