A ball is thrown straight upward and returns
to the thrower’s hand after 2.3 s in the air. A
second ball is thrown at an angle of 60◦ with
the horizontal.
At what speed must the second ball be
thrown so that it reaches the same maximum
height as the first? The acceleration of gravity
is 9.8 m/s
2
.
Answer in units of m/s
To find the speed at which the second ball must be thrown in order to reach the same maximum height as the first ball, we need to consider the kinematic equations for projectile motion.
For the first ball:
We know that the total time it takes for the ball to return to the thrower's hand is 2.3 seconds. Since the ball is thrown straight upward, the time taken to reach the maximum height will be half of this total time, which is 2.3/2 = 1.15 seconds.
Using the kinematic equation for vertical displacement:
y = v0y * t - (1/2) * g * t^2
Where:
y is the vertical displacement or maximum height (which is the same for both balls)
v0y is the initial vertical velocity component
g is the acceleration due to gravity (9.8 m/s^2)
t is the time taken to reach the maximum height (1.15 seconds for the first ball)
Since the first ball is thrown straight upward, the initial vertical velocity component v0y will be equal to the final vertical velocity component. So, we can rewrite the equation as:
y = vfy * t - (1/2) * g * t^2
Since the ball is at its maximum height, the final vertical velocity component, vfy, will be zero. Therefore, the equation becomes:
y = - (1/2) * g * t^2
Now, let's calculate the maximum height that the first ball reaches. Plug in the given values:
y = - (1/2) * 9.8 * (1.15)^2
Solving this equation, we get:
y = -6.647625 m
Now, to find the speed at which the second ball must be thrown, we can use the range equation for projectile motion:
R = v0^2 * sin(2θ) / g
Where:
R is the range or horizontal displacement (which is zero for the second ball, as it reaches the same maximum height as the first ball)
v0 is the initial velocity of the second ball
θ is the launch angle (60° in this case)
g is the acceleration due to gravity (9.8 m/s^2)
Since the range is zero for the second ball, the equation becomes:
0 = v0^2 * sin(2 * 60°) / 9.8
Now, let's solve for v0:
v0^2 * sin(120°) = 0
Since sin(120°) is positive, we can ignore it. Therefore,:
v0^2 = 0
The only possible solution is v0 = 0.
So, the second ball must be thrown with a speed of 0 m/s in order to reach the same maximum height as the first ball.
To find the speed at which the second ball must be thrown to reach the same maximum height as the first ball, we can use the equations of motion under constant acceleration.
Let's consider the motion of each ball separately:
For the first ball:
- The time taken to reach the maximum height is half of the total time in the air, which is 2.3 s / 2 = 1.15 s.
- The acceleration of gravity acts downward throughout the motion, so we can use -9.8 m/s^2 as the value for acceleration.
- The initial velocity of the ball is the same as the final velocity when it returns to the thrower's hand, which is 0 m/s at the maximum height.
Using the equation of motion for vertical motion, we have:
Vf = Vi + at, where
- Vf is the final velocity (0 m/s at the maximum height),
- Vi is the initial velocity (which we need to find),
- a is the acceleration (-9.8 m/s^2), and
- t is the time taken to reach the maximum height (1.15 s).
0 = Vi + (-9.8) * (1.15)
Vi = 9.8 * 1.15
Vi = 11.27 m/s
So, the initial velocity of the first ball is 11.27 m/s.
For the second ball:
- The motion is at an angle of 60 degrees with the horizontal.
- To compare the vertical motion with the first ball, we need to consider the vertical component of the velocity.
- The vertical component can be found using the equation Vy = V * sin(theta), where
- Vy is the vertical component of the velocity,
- V is the velocity we need to find, and
- theta is the angle of 60 degrees.
So, Vy = V * sin(60).
At the maximum height, the vertical component of the second ball's velocity will be the same as that of the first ball. Thus, we have:
Vy (second ball) = Vy (first ball)
V * sin(60) = 11.27 m/s
Simplifying this equation, we get:
V = 11.27 m/s / sin(60)
V ≈ 13.01 m/s
Therefore, the second ball must be thrown at a speed of approximately 13.01 m/s to reach the same maximum height as the first ball.
2.3/2 = 1.15 seconds to go up
0 = Vi -4.9 t = Vi - 4.9*1.15
Vi = 5.365 m/s initial speed up
the second ball must have same initial vertical component of velocity = 5.365 m/s
Vi sin 60 = 5.365
so
Vi = 6.51 m/s