Total time In air for a projectile

[bionut]

1. A springboard diver performs a dive from a 10 metre tower and jumps with an initial vertical velocity of 5 ms and horizontal velocity of 2 ms. Determine the time she has in the air to perform her dive?

The Attempt at a Solution

From what I know 0=Vi + at (will give me 1/2 the flight time)
t=5/9.81
t=-0.51 X 2 = Total = 1.02 s... whhich is wrong (do I need to incorperate the Disatnce) or should I try and find the angle between the 2 velocities?

 

[PeterO]

1. A springboard diver performs a dive from a 10 metre tower and jumps with an initial vertical velocity of 5 ms and horizontal velocity of 2 ms. Determine the time she has in the air to perform her dive?

The Attempt at a Solution

From what I know 0=Vi + at (will give me 1/2 the flight time)
t=5/9.81
t=-0.51 X 2 = Total = 1.02 s... whhich is wrong (do I need to incorperate the Disatnce) or should I try and find the angle between the 2 velocities?

You have calculated how long until the diver passes the tower on the way down. She still has to get all the way to the water!

 

[bionut]

would it just be as siple as 10m height @ 5m/s = 2s?

 

[PeterO]

would it just be as siple as 10m height @ 5m/s = 2s?

Unfortunately - NO.

The diver will be accelerating all the way down.

If you use one of the displace ment formulae, rather than just the velocity formula, the answer should come out.

Like for example X = V_ot + 0.5 a t²

make sure you are consistent with positive and negative.
Decide which direction [up or down] will be positive, and be consistent with the displacement, Initial velocity and acceleration.

 

[rwisz]

I would approach this problem by first breaking down the motion of the diver into its vertical and horizontal components. The initial vertical velocity of 5 m/s can be used to calculate the time the diver spends in the air using the equation d = Vit + 1/2at^2. Since the initial vertical velocity is 5 m/s and the acceleration due to gravity is -9.81 m/s^2, the equation becomes 0 = 5t - 4.905t^2. Solving for t gives us t = 1.02 seconds, which is the time the diver spends in the air.

To determine the total time in the air, we also need to consider the horizontal motion of the diver. Since there is no acceleration in the horizontal direction, the horizontal velocity of 2 m/s remains constant throughout the motion. Therefore, the total horizontal distance traveled by the diver is given by d = Vxt, where Vx is the horizontal velocity and t is the time spent in the air. Substituting in the values, we get d = 2 x 1.02 = 2.04 meters.

Therefore, the total time in the air for the projectile is 1.02 seconds, and the total horizontal distance traveled is 2.04 meters. This information can be used to accurately predict the trajectory of the diver and ensure a safe and successful dive.