A person who is 1.1 m tall throws a ball with a speed of 14 m/s at an angle of 71 degrees above the horizontal, as shown in the diagram http://i49.tinypic.com/vdmfb7.png
h = 1.1 m, v = 14 m/s, θ = 71 degrees
Use the acceleration due to gravity as g = 10 m/s2.
What is the time of flight; that is, the time taken for the ball to hit the ground after it has left the person's hand?
Please show step-by-step. I want to learn! AND I PROMISE 10 POINTS FOR YOUR TIME AND EFFORT!
h = 1.1 m, v = 14 m/s, θ = 71 degrees
Use the acceleration due to gravity as g = 10 m/s2.
What is the time of flight; that is, the time taken for the ball to hit the ground after it has left the person's hand?
Please show step-by-step. I want to learn! AND I PROMISE 10 POINTS FOR YOUR TIME AND EFFORT!
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The trajectory of the ball is symmetric about its max height above its release height. Once at max height, the ball falls an additional 1.1 m = h, to the ground.
Vo = 14 m/s at 71°
Voy = 14 sin 71° = 13.2 m/s = initial vertical velocity
time to reach max height = Voy/g = 13.2/10 = 1.32 s
max height above release point = 1/2gt² = (0.5)(10)(1.32)² = 8.71 m
vertical distance from max height to ground = 8.71 + h + 8.71 + 1.1 = 9.81 m = H
time to fall = T, from max height (where initial vertical velocity = 0) to ground
H = 1/2gT²
9.81 = (0.5)(10)T²
T² = 9.81/(0.5)(10) = 9.81/5 = 1.962
T = √1.962 = 1.40 s
TOTAL TIME of flight of ball = t + T = 1.32 + 1.40 = 2.7 s ANS
Vo = 14 m/s at 71°
Voy = 14 sin 71° = 13.2 m/s = initial vertical velocity
time to reach max height = Voy/g = 13.2/10 = 1.32 s
max height above release point = 1/2gt² = (0.5)(10)(1.32)² = 8.71 m
vertical distance from max height to ground = 8.71 + h + 8.71 + 1.1 = 9.81 m = H
time to fall = T, from max height (where initial vertical velocity = 0) to ground
H = 1/2gT²
9.81 = (0.5)(10)T²
T² = 9.81/(0.5)(10) = 9.81/5 = 1.962
T = √1.962 = 1.40 s
TOTAL TIME of flight of ball = t + T = 1.32 + 1.40 = 2.7 s ANS
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X Velocity = COS 71 x 14 m/s = 4.56 m/s
Y Velocity = SIN 71 x 14 = 13.24 m/s
Y travel up = 13.24 / 10 = 1.324 seconds
Y distance up = 1/2 AT^2 = 5 x 1.324^2 = 8.76 meters
Y distance down = 8.76 + 1.1 = 9.86 meters
9.86 = 1/2 10 T^2 = SQRT 9.86 / 5 = 1.404 seconds
Total time = 1.404 +1.324 = 2.728 seconds (answer)
X distance traveled = 4.56 m/s times 2.728 seconds = 12.44 meters (answer)
Y Velocity = SIN 71 x 14 = 13.24 m/s
Y travel up = 13.24 / 10 = 1.324 seconds
Y distance up = 1/2 AT^2 = 5 x 1.324^2 = 8.76 meters
Y distance down = 8.76 + 1.1 = 9.86 meters
9.86 = 1/2 10 T^2 = SQRT 9.86 / 5 = 1.404 seconds
Total time = 1.404 +1.324 = 2.728 seconds (answer)
X distance traveled = 4.56 m/s times 2.728 seconds = 12.44 meters (answer)
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This sort of problem is fun! You need to find how fast the ball starts to rise when first thrown. That is called the "vertical component" of the ball's motion. In your case, the ball is travelling, overall, at 14 m/s. Its vertical component is 14 * Sine 71 degrees. That is 14 m/s * .9455 = 13.24 m/s. That is its initial vertical velocity. With an acceleration of gravity of 10 m/s/s. it will rise for 1.3 seconds.
How much will it rise from its starting elevation of 1.1m? Well, its average velocity is (13.24 + 0)/2. That's 6.62m/s for an average speed... and, in 1.3 seconds it will rise 8.6 meters.
Since it started out at 1.1m its final height will be 8.6 + 1.1 = 9.7m. And now it has to fall back to the earth from its height of 9.7m.
Distance = 1/2 accel. * t^2; 9.7m = 1/2(10)*t^2; 9.7 = 5t^2; 1.94 = t^2; t = 1.39 seconds.
So, it rises for 1.3 seconds, falls for 1.4 seconds... a total of 2.7 seconds.
I know yow didn't ask....
How far did it travel horizontally? The horizontal component of the ball's velocity is the Cosine of 71 degrees * 14 m/s... about 4.6 m/s. At that rate, in 2.7 seconds it will hit the ground 12.4 m from the thrower.
How much will it rise from its starting elevation of 1.1m? Well, its average velocity is (13.24 + 0)/2. That's 6.62m/s for an average speed... and, in 1.3 seconds it will rise 8.6 meters.
Since it started out at 1.1m its final height will be 8.6 + 1.1 = 9.7m. And now it has to fall back to the earth from its height of 9.7m.
Distance = 1/2 accel. * t^2; 9.7m = 1/2(10)*t^2; 9.7 = 5t^2; 1.94 = t^2; t = 1.39 seconds.
So, it rises for 1.3 seconds, falls for 1.4 seconds... a total of 2.7 seconds.
I know yow didn't ask....
How far did it travel horizontally? The horizontal component of the ball's velocity is the Cosine of 71 degrees * 14 m/s... about 4.6 m/s. At that rate, in 2.7 seconds it will hit the ground 12.4 m from the thrower.