Help Projectile Motion- Firing of a gun

[fewill]

1. A projectile is fired from a gun (on the ground)and has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively.a) How long does it take the projectile to reach the highest point on the trajectory? b) What is the magnitude of the projectile's velocity just before striking the ground? c) How far does the projectile travel horizontally before it hits the ground?

Homework Equations

The Attempt at a Solution

a) t= vy-viy/g 0-40/9.8= 4.08s
b) v= square root of viy^2 + vix^2= 50
v=viy+gt 0-9.8*4.08=40 m/s
c) total time = 2*4.08, 30*8.16= 244.8 m

 

[HallsofIvy]

1. A projectile is fired from a gun (on the ground)and has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively.a) How long does it take the projectile to reach the highest point on the trajectory? b) What is the magnitude of the projectile's velocity just before striking the ground? c) How far does the projectile travel horizontally before it hits the ground?

Homework Equations

The Attempt at a Solution

a) t= vy-viy/g 0-40/9.8= 4.08s
b) v= square root of viy^2 + vix^2= 50
v=viy+gt 0-9.8*4.08=40 m/s
c) total time = 2*4.08, 30*8.16= 244.8 m

You and reverse should get together!

(a) Yes, at maximum height the y component of velocity is 0 and, at acceleration g, it will take (vy- viy)/g seconds to reach that height. (I don't know why you switched to 0-40 instead of 40-0 obviously the time is NOT negative.)

(b) You've actually calculated the speed (magnitude of velocity) initially but, yes, because of the symmetry that is the same as the speed at the end of the trajectory.

(c) I don't know why you wrote "v=viy+gt 0-9.8*4.08=40 m/s". Obviously, 0-9.8*4.08= -40 m/s, not 40 m/s, but in any case, that is the vertical component of speed just as the projectile hits the ground- which you were not asked. Yes, again because of the symmetry of the trajectory, the total time is exactly twice the time to maximum height and the horizontal distance is just the horizontal component of velocity multiplied by that time.

 

[baba89]

I would first like to clarify that the given information is not enough to accurately solve for the time, velocity, and distance traveled by the projectile. The equations used in the attempt at a solution are correct, but without knowing the angle at which the projectile was fired, we cannot accurately calculate these values. The angle of projection is a crucial factor in determining the trajectory of the projectile and its velocity at different points in its motion. Additionally, the given values for the initial horizontal and vertical components of velocity may not be realistic for a gun fired on the ground. In a real-life scenario, the horizontal component of velocity would likely be much smaller than the vertical component.

Therefore, in order to accurately solve for the time, velocity, and distance traveled by the projectile, we would need to know the angle of projection, as well as any other relevant factors such as air resistance and the mass of the projectile. With this additional information, we can use the equations for projectile motion to calculate the values requested in the question.

Furthermore, it is important to note that in a real-life scenario, the trajectory of the projectile would also be affected by external factors such as wind, air resistance, and the rotation of the Earth. These factors would need to be taken into account in order to accurately predict the motion of the projectile.

In conclusion, while the attempt at a solution is correct based on the given information, as a scientist, I would recommend gathering more data and considering external factors in order to accurately solve for the time, velocity, and distance traveled by the projectile.