Projectile Motion Problem: Calculating Velocity, Distance, and Potential Energy

AI Thread Summary
The discussion centers on solving a projectile motion problem involving a 45 kg projectile launched at a 30° angle, which remains in the air for 48 seconds. Participants emphasize that the time to reach maximum height is half the total time of flight, allowing for the calculation of initial vertical velocity using gravitational acceleration. The initial horizontal velocity can then be derived from the vertical velocity, leading to the total initial velocity and horizontal distance traveled. The maximum height is determined through basic kinematics, utilizing known time and acceleration values. Understanding that gravity is the sole force acting on the projectile is crucial for these calculations.
Flanery
Messages
2
Reaction score
0

Homework Statement


"A 45.0 kg projectile is fired from a gun. The elevation angle of the gun is 30.0°. the projectile is in the air for 48.0 s before it hits the deck at the same height as it was fired. Ignoring friction find: the projectile's initial velocity, the horizontal distance the projectile traveled, and the potential energy of the projectile at it's highest point in its trajectory.

Weight: 45 kg
Angle of launch: 30°
Total time in air: 48 s

Homework Equations


D = VT

The Attempt at a Solution


We haven't done projectile motion problems since the beginning of the year. It seems this problem is a random one on the homework assignments. I've completely forgotten how to calculate these. I drew out the motion of it but it's not ringing any bells.
 
Physics news on Phys.org
Welcome to PF!

By symmetry, you know that the time it takes to reach the max height is half of the total travel time (think about it if you're not sure).

Since you know the vertical acceleration, and the time taken to reach the max height (i.e. the time needed to reduce the vertical velocity to zero), you can determine the initial vertical velocity.

Since you know the angle of launch, you can use the above result to determine the initial horizontal velocity, and from that, the total initial velocity.

Once you know the horizontal speed, you can figure out the total horizontal distance travelled.

The max height comes from basic kinematics. You already know the time taken to reach it, and the acceleration.

I hope this helps.
 
cepheid said:
Since you know the vertical acceleration, and the time taken to reach the max height (i.e. the time needed to reduce the vertical velocity to zero), you can determine the initial vertical velocity.

Hey, thanks for your response! I don't quite understand how I have the vertical acceleration. Do I need to find it from the weight/gravity/time?
 
Flanery said:
Hey, thanks for your response! I don't quite understand how I have the vertical acceleration. Do I need to find it from the weight/gravity/time?

What defines projectile motion is that the particle is solely under the influence of gravity. That's the only force that acts on it. So, the vertical acceleration is due to gravity.
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top