A ball is thrown straight upward and returns to the thrower's hand after 2.00 s in the air. A second ball thrown at an angle of 40.0° with the horizontal reaches the same maximum height as the first ball.
(a) At what speed was the first ball thrown?
m/s
(b) At what speed was the second ball thrown?
m/s
To solve this problem, we can use the kinematic equations of motion. Let's break it down step by step.
(a) At what speed was the first ball thrown?
We know that the time of flight of the first ball, which is the time taken for it to go up and then come back down, is 2.00 seconds.
Using the kinematic equation for vertical motion:
Δy = Viy * t + (1/2) * g * t²
Where:
Δy is the vertical displacement or change in height
Viy is the initial vertical velocity
t is the time of flight
g is the acceleration due to gravity (approximately 9.8 m/s²)
Since the ball returns to the thrower's hand, the vertical displacement is zero.
0 = Viy * t + (1/2) * g * t²
Simplifying the equation, we have:
0 = Viy * 2 + (1/2) * 9.8 * (2)²
0 = 2Viy + 19.6
Rearranging the equation:
2Viy = -19.6
Viy = -9.8 m/s
The negative sign indicates that the initial vertical velocity is in the opposite direction of the acceleration due to gravity.
The initial vertical velocity (Viy) is equal to the magnitude of the final vertical velocity when the ball reaches its maximum height. So, the speed at which the first ball was thrown is 9.8 m/s.
(b) At what speed was the second ball thrown?
To find the speed at which the second ball was thrown, we need to split the initial velocity into its horizontal and vertical components.
Vix = Vi * cosθ
Viy = Vi * sinθ
Where:
Vix is the initial horizontal velocity
Viy is the initial vertical velocity
Vi is the initial velocity of the second ball
θ is the angle of projection (40.0° in this case)
Since the maximum height reached by the second ball is the same as the first ball, the vertical displacement is also zero.
0 = Viy * t + (1/2) * g * t²
0 = Vi * sinθ * t + (1/2) * 9.8 * t²
Dividing both sides by t, we get:
0 = Vi * sinθ + (1/2) * 9.8 * t
Since sinθ = sin(40.0°) = 0.6428 and t = 2.00s, we can substitute these values into the equation:
0 = Vi * 0.6428 + (1/2) * 9.8 * 2.00
Simplifying the equation:
0 = 0.6428Vi + 9.8
Rearranging the equation:
0.6428Vi = -9.8
Vi = -15.23 m/s
Again, the negative sign indicates the opposite direction of the acceleration due to gravity.
Therefore, the speed at which the second ball was thrown is 15.23 m/s.
Tr = Rise time, Tf = Fall time.
a. Tr + Tf = 2 s, Tf = Tr, Tr+Tr = 2, 2Tr = 2, Tr = 1 s.
V = Vo + g*Tr = 0, Vo = -g*Tr =
9.8*1 = 9.8 m/s.
b. Yo = 9.8 m/s(Part a), Vo*sin40
= 9.8, Vo = 9.8/sin40 = 15.25 m/s.