need help with physics homework on trajectories and gravity plz?
Students in lab measure the speed of a steel ball launched horizontally from a table top to be 4.1 m/s. If the table top is 1.6 m above the floor, where should they place a 20 cm tall tin coffee can to catch the ball when it lands?
First you want to determine the time it takes for the steel ball to travel from 1.6 m above the floor to 0.2 m (20 cm) above the floor (the 0.2 m above the floor equals the height of the coffee can). The height of the coffee can above the floor matters because you want to place the height of the can at the instant the ball reaches the height of the can.
Finally you can determine how far away from the edge of the table top you can place the coffee can.
1). Determine the time if takes for the ball to reach the same height as the coffee can:
Δy = Voyt + 0.5*ay*t^2
where:
Δy = vertical distance from the table top to the top of the can (1.6 m – 0.2 m = 1.4 m)
Voy = initial velocity of the ball in the vertical (0 m/s, the ball is launched horizontally)
t = time it takes for the ball to reach the height of the can
ay = acceleration due to gravity (9.81 m/s^2)
Δy = Voyt + 0.5*ay*t^2
Δy = 0 + 0.5*ay*t^2
Solving for t,
t = √(2*Δy/ay)
t = √(2*1.4/9.81)
t = 0.53425 seconds
2). Determine the horizontal distance from the table top to the coffee can:
Δx = Voxt + 0.5*ax*t^2
where:
Δx = horizontal distance from the table top to the coffee can
Vox = initial velocity of the ball in the horizontal (4.1 m/s)
ax = horizontal gravitational acceleration (0 m/s, no gravity in the horizontal)
t = time for the ball to reach the height of the coffee can (0.53425 seconds)
Δx = Voxt + 0.5*ax*t^2
Δx = Voxt
Δx = (4.1 * 0.53425)
==>Δx = 2.19 m
I hope this helped!
First you want to determine the time it takes for the steel ball to travel from 1.6 m above the floor to 0.2 m (20 cm) above the floor (the 0.2 m above the floor equals the height of the coffee can). The height of the coffee can above the floor matters because you want to place the height of the can at the instant the ball reaches the height of the can.
Finally you can determine how far away from the edge of the table top you can place the coffee can.
1). Determine the time if takes for the ball to reach the same height as the coffee can:
Δy = Voyt + 0.5*ay*t^2
where:
Δy = vertical distance from the table top to the top of the can (1.6 m – 0.2 m = 1.4 m)
Voy = initial velocity of the ball in the vertical (0 m/s, the ball is launched horizontally)
t = time it takes for the ball to reach the height of the can
ay = acceleration due to gravity (9.81 m/s^2)
Δy = Voyt + 0.5*ay*t^2
Δy = 0 + 0.5*ay*t^2
Solving for t,
t = √(2*Δy/ay)
t = √(2*1.4/9.81)
t = 0.53425 seconds
2). Determine the horizontal distance from the table top to the coffee can:
Δx = Voxt + 0.5*ax*t^2
where:
Δx = horizontal distance from the table top to the coffee can
Vox = initial velocity of the ball in the horizontal (4.1 m/s)
ax = horizontal gravitational acceleration (0 m/s, no gravity in the horizontal)
t = time for the ball to reach the height of the coffee can (0.53425 seconds)
Δx = Voxt + 0.5*ax*t^2
Δx = Voxt
Δx = (4.1 * 0.53425)
==>Δx = 2.19 m
I hope this helped!
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