A ball is thrown straight upward and returns to the thrower's hand after 2.30 s in the air. A second ball is thrown at an angle of 42.0° with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically?
damon how you go from sin 42 to sin 32
first ball stops in 2.3/2 = 1.15 seconds
0 = Vi - g t
Vi = 9.81 (1.15) = 11.3 m/s
second ball has same Vi
Vi = V sin 42
V = 11.3/sin 32 = 21.3 m/s
To find the speed at which the second ball must be thrown, we can use the principles of projectile motion. Here's how you can solve it step-by-step:
Step 1: Determine the vertical displacement of the vertically thrown ball.
Since the ball is thrown straight up and returns to the thrower's hand, the vertical displacement is zero. This means the two balls will reach the same maximum height.
Step 2: Find the time it takes for the second ball to reach the same height.
The time of flight for the vertically thrown ball is 2.30 s. Since the second ball has the same maximum height, the time it takes to reach that height will also be 2.30 s.
Step 3: Split the initial velocity of the second ball into horizontal and vertical components.
The initial velocity can be split into two components: one parallel to the horizontal direction (Vx) and one parallel to the vertical direction (Vy). We can calculate these components using the given angle of 42.0° and the magnitude of the velocity.
Vx = V * cos(θ)
Vy = V * sin(θ)
Step 4: Determine the vertical displacement of the second ball.
To reach the same height as the vertically thrown ball, the vertical displacement of the second ball must be zero (since both balls reach the same maximum height). Use the following equation to calculate the vertical displacement:
y = (Vy * t) + (0.5 * g * t^2)
Since the ball reaches the same height within the same time, we can use the time of flight (2.30 s) instead of t.
0 = (Vy * 2.30) + (0.5 * g * 2.30^2)
Step 5: Solve for the magnitude of the velocity (V).
Rearrange the equation from Step 4 to solve for V:
V = -((0.5 * g * t^2) / t)
Substituting the values of g (acceleration due to gravity, approximately 9.8 m/s²) and t (2.30 s) into the equation, we can calculate the required velocity.
V = -((0.5 * 9.8 * 2.30^2) / 2.30)
Simplifying the equation, we get:
V ≈ 10.8 m/s
Therefore, the second ball must be thrown at approximately 10.8 m/s for it to reach the same height as the vertically thrown ball.