Motion in two dimensions basketball dunk

[ppxrare]

Homework Statement

A basketball star covers 2.40 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled precisely as that of a particle at his center of mass. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.90 m above the floor and is at elevation 0.910 m when he touches down again.

(a) Determine his time of flight (his "hang time").


(b) Determine his horizontal velocity at the instant of takeoff.

(c) Determine his vertical velocity at the instant of takeoff.


(d) Determine his takeoff angle.


(e) For comparison, determine the hang time of a whitetail deer making a jump (see figure above) with center-of-mass elevations yi = 1.20 m, ymax = 2.45 m, and yf = 0.800 m.


I tried to think about the question but I'm not getting anywhere, by time of flight do they mean time till he reaches max height or till he reaches hoop and comes down again and I am confused by the elevation

 

[p21bass]

Time of flight = total time from when he leaves the ground to when he comes back down. As far as the elevation is concerned, you simply have two different values of delta-y. So, you need to fully calculate the time to go from 1.02m to 1.90m, then from 1.90m to 0.91m and add those two together. You'll do the same with the deer in part e.

 

[ppxrare]

how do I calculate the time though? I don't think I have enough information as I don't have initial velocity

 

[p21bass]

No - you're supposed to calculate initial velocity for parts b & c. You do have equations that relate distance, time, acceleration (hint #1: what is the only acceleration involved?), initial velocity, and final velocity (hint #2: what is the velocity in the y-direction at the top of his arc?).

 

[ppxrare]

ok so far this is what I have thought of, delta y = 0.88m final y velocity is 0 (top of arc) and acceleration is -9.81

I used these two equations

delta y = Viyt -9.81t^2
Vyf= Vyi + at

I substituted for Viy but I got the time negative what am I doing wrong?

 

[p21bass]

Your delta-y and finaly y velocity are correct, and the second equation is correct. And you obviously know that a is -9.81m/s^2. But, your first equation is slightly off. It is 1/2*at^2 - you forgot the 1/2.

But, I think you need to first use the equation that relates the final and initial velocities to the distance and acceleration. Since you already have three of those variables (a=-9.81m/s^2, delta-y=0.88m, and v_yf=0), it should be easy to find v_yi and then plug that value into one of your equations to find the first t.

 

[ppxrare]

thanks a lot man, I got it correct. Would appreciate some more help in another thread:P