# Kinematics Questions

1. Jul 11, 2014

### kathy

I'm taking grade 11 physics online for the summer time. I have a problem with this question :] An airplane flying at 50.0m/s is bringing food and emergency first aid supplies to a camp. The plane is for the pilot to drop the supplies so that they land on an "X" marked on the ground 150m below. How long will it take the supplies to fall to the ground? How far in front of the "X" should the pilot release the supplies so that they land directly on the "X?"

I just want to know the formula that I should use. Sorry. I'm just really confused. Physics is my worst subject but I wanted to get it over with during the summer so that I could have some room for other courses during the year.

2. Jul 11, 2014

### Nick O

Well, I can hint at the formulas (you need two different ones!) you need.

1. The package will continue to move forward at 50.0 m/s. The forward velocity is constant.

2. The package will accelerate downward at g. Your textbook probably defines g as either 9.8 m/s^2 or 10 m/s^2. The downward acceleration (but not velocity) is constant.

3. Jul 11, 2014

### Staff: Mentor

Welcome to the PF Kathy.

Per the PF rules (please see Site Info at the top of the page), you need to show your work in order for us to provide tutorial help. Please use the hints you've been given to start working out the math of the solution. Please post your work so that we can help you.

Please also in the future, use the Homework Help Template that you are provided when posting a new thread in the HH forums. The Template will help you to organize your thoughts, and will help us to help you.

Thank you.

4. Jul 13, 2014

### kathy

Problem Solving

Sorry I didn't problem solve before but I'll make sure to the next time I post a thread.
Here's my problem solving.
a. How long will it take the supplies to fall to the ground? (You can ignore the effect of air resistance).
d= -150m
v1= 50m/s
a= 9.8m/s2
d = (v) (t) + (a) (t2) /2
t= (2) (d) / a
t= square root of (2) (-150) / -9.8m/s2
t= square root of -300/ -9.8m/s2
t= square root of 30.6122449
t= 5.53s
v2= v1+ (a) (t)
v2= 0 + (-9.8m/s2) (5.53s)
v2= -54.194
v2= -54.2m/s
Therefore it will take the supplies approximately 5.53s to hit the "X" marked on the ground 150m below.
b. How far in front of the "X" should the pilot release the supplies so that they land directly on the "X"?
t= 5.53s
v1= 50m/s
acceleration = 0m/s2
displacement = ?
d= (v) (t)
d= (50m/s)(5.53s)
d= 276.5
d= 276.5
Therefore in order for the pilot to release the supplies so that they land directly on the "X" the pilot should release it about 276.5 m away from the "X".

I think I messed up with part b) but I'm alright with part a).

5. Jul 13, 2014

### Satvik Pandey

I think your part a is not correct ,initial velocity in y direction is zero.The given velocity is velocity of particle in x direction .This is a problem of projectile motion(Motion in two dimensions).
Try to find time required to hit the ground(take u=0 in part a ) and multiply this time with horizontal velocity to get answer.

Last edited: Jul 13, 2014