1. Harry accidentally falls out of a helicopter that is traveling horizontally at 100 m/s. He plunges into the ocean below him 2.2 seconds later. How high above the ocean was the helicopter when Harry fell out?
Hiw far did Harry travel horizontally while falling?
3. Nicole is standing in a 20 m tall tower and throws a ball that travels 60 m horizontally across the ground. Assuming the ball is thrown in a perfectly horizontal direction, at what velocity did Nicole throw the ball?
4. Through "His Airness," Michael Jordan, was best known for having an incredible "hang time" when he jumped, 5'8" Spud Webb was actually the only player in the NBA who could stay in the air for longer than 1 second. If Spud stays in the air for 1 sec, how high did he jump?
I will be glad to critique your work.
They all can be solved by using the equation for distance fallen vs time.
You should know that equation by now.
I got -23.716 m and 220 m for the first one, the second I don't know how to do, and the third i got -4.9 m
i just want to make sure those answers are right.
For problem #3, calculate how long it takes to fall 20 m.
60 m divided by that time will be the initial horizontal speed.
Telling us what you got is not the same as showing your work.
Your last answer is obviously wrong. A negative distance is not a jump, and no one can jump 4.9 meters.
Don't listen to drwls for #4. The answer is 5m/2. this is because you travel 5m in 1 second. So you jump and fall down. So it is 5m/2 and the answer is 2.5 m/s.
To answer these questions, we can use the equations of motion and the principles of projectile motion. Here are the steps to solve each problem:
1. To find the height of the helicopter, we can use the equation for vertical displacement:
Δy = v₀t + (1/2)at²
In this case, the initial vertical velocity (v₀) is 0 m/s because Harry fell out of the helicopter. The acceleration (a) due to gravity is 9.8 m/s² (assuming no air resistance). The time (t) is given as 2.2 seconds.
Therefore, Δy = 0(2.2) + (1/2)(9.8)(2.2)² = 24.86 meters
Harry fell from a height of approximately 24.86 meters above the ocean.
2. To find the horizontal distance traveled by Harry, we can use the equation for horizontal displacement:
Δx = vxt
In this case, the horizontal velocity (vx) is given as 100 m/s, and the time (t) is again 2.2 seconds.
Therefore, Δx = 100(2.2) = 220 meters
Harry traveled a horizontal distance of 220 meters while falling.
3. To find the initial velocity of the ball thrown by Nicole, we can use the equation for horizontal displacement:
Δx = vxt
In this case, the horizontal displacement (Δx) is given as 60 meters, and the time (t) is not provided.
Therefore, we need to first find the time it took for the ball to travel the given horizontal distance. We can use the equation for horizontal velocity:
vx = Δx / t
Rearranging the equation, we have:
t = Δx / vx
t = 60 / vx
Since the ball is thrown horizontally, there is no vertical displacement, and the time of flight (t) is the same as the time it took to travel the horizontal distance.
Now, to calculate the initial velocity (v₀), we can use the equation for vertical displacement:
Δy = v₀t + (1/2)at²
In this case, the vertical displacement (Δy) is given as the height of the tower, 20 meters. The acceleration (a) due to gravity is 9.8 m/s², and we found the time of flight (t) to be 60 / vx.
Therefore, 20 = v₀(60 / vx) + (1/2)(9.8)(60 / vx)²
Rearranging and solving for v₀, we get:
v₀ = (20 - (1/2)(9.8)(60 / vx)²) * (vx / (60 / vx))
Simplifying further, we have:
v₀ = (20 * vx) / (1 + (1/2)(9.8)(60 / vx)²)
Substitute the given value of vx to calculate the initial velocity.
4. To find the height jumped by Spud Webb, we can use the equation for vertical displacement:
Δy = v₀t + (1/2)at²
In this case, the initial vertical velocity (v₀) is given as 0 m/s (assuming he jumped vertically). The acceleration (a) due to gravity is 9.8 m/s², and the time (t) is given as 1 second.
Therefore, Δy = 0(1) + (1/2)(9.8)(1)² = 4.9 meters
Spud jumped to a height of 4.9 meters.