A bullet is fired from ground level with a speed of 75.0m/s...find velocity at the highest point of trajectory

Question

A bullet is fired form ground level with a speed of 75.0m/s at an angle of 30.0 degrees above the horizontal at a location where g=10m/s^2. What is the horizontal component of its velocity when it is at the highest point of its trajectory?

Can you please show me how to do this.

Answer 1

For this one, use a bit of trig to separate the horizontal and vertical components of velocity.

At the apogee, the vertical component will be zero, so you're only left with the horizontal:

vh=75*cos(30)

j

Answer 2

i dont really know what u r asking but this is a projectile motion question for sure.

MY FIRST GUESS IS cos(30)(75)=64.95m/s as the horizontal velocity of a projectile is always constant.

i can help you though time is constant in horizontal and vertical dispplacements. so you can find the time dx=v^2*Sin2(theta)/g
therefore dx={cos(30)(75)}^2*Sin(60)/10
dx=365.35 m (the horizontal displacement)

so for time t = 365.35/cos(30)(75)
t=5.62s

and if u divide this time by 2 you will get 2.81s which will represent the axis of symmetry for the parabolic arc , at this time the bullet will be at its peak. Since i didn;t understand your question properly i dont know what to say so i gave you all the unknowns. You can use motion formulas etc to find for any other unknown if you want to approach the problem in another way but what i personally believe is that horizontal component of the velocity is always the same as its constant so it should be 64.95m/s. But like i said i didn;t understand ur question so thats all i could do.