A ball is thrown straight upward and returns to the thrower's hand after 2.75 s in the air. A second ball thrown at an angle of 34.0° 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
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To solve this problem, we can use the kinematic equations of motion. Let's start by analyzing the motion of the first ball.
(a) To find the initial speed of the first ball (v₀), we can use the equation for displacement (Δy) in vertical motion:
Δy = v₀t + (1/2)gt²
In this equation, Δy represents the change in height, v₀ is the initial speed, t is the time the ball is in the air, and g is the acceleration due to gravity (approximately 9.8 m/s²).
Since the ball is thrown straight upward and returns to the thrower's hand, the change in height, Δy, is zero. The time it takes for the ball to reach its maximum height and return back down is 2.75 seconds.
Therefore, we can rewrite the equation as:
0 = v₀(2.75) + (1/2)(9.8)(2.75)²
Simplifying the equation gives:
0 = v₀(2.75) + 17.13625
Now we can solve for v₀:
v₀(2.75) = -17.13625
v₀ = -17.13625 / 2.75
v₀ ≈ -6.23 m/s
Since the speed cannot be negative, we take the absolute value to obtain the magnitude:
v₀ ≈ 6.23 m/s
Therefore, the first ball was thrown with a speed of approximately 6.23 m/s.
(b) Now let's find the speed at which the second ball was thrown. Since it is thrown at an angle of 34.0° with the horizontal, we can decompose its motion into vertical and horizontal components.
The maximum height reached by the second ball is the same as that of the first ball. Therefore, we can use the same equation in the vertical direction:
0 = v₀sinθt + (1/2)gt²
θ represents the angle of projection, v₀ is the initial speed, t is the time of flight, and g is the acceleration due to gravity.
The time of flight will be the same as that of the first ball, which is 2.75 seconds.
From the equation, we can rearrange to solve for v₀:
v₀sinθt = -17.13625
v₀sin34°(2.75) = -17.13625
Now we can solve for v₀:
v₀ ≈ -17.13625 / (sin34° * 2.75)
v₀ ≈ -13.47 m/s
Again, we take the magnitude to obtain the speed:
v₀ ≈ 13.47 m/s
Therefore, the second ball was thrown with a speed of approximately 13.47 m/s.