A projectile is fired straight upward at 141 m/s.?

Question

How fast is it moving at the instant it reaches the top of its trajectory? Suppose instead that it were fired upaward at 45 degrees. What would be its speed at the top of its trajectory

Answer 1

At the top of a particle's trajectory, its upward velocity is zero.

If the particle only originally had upward velocity, then at the top of the trajectory its total velocity would be zero.
If the particle is fired at an angle it has both a horizontal and vertical velocity component. At the top of its trajectory the vertical component is zero but the horizontal component is the same as it was as it left the ground.

vertical component of velocity = v * sin (theta)
horizontal component of velocity = v * cos (theta)
where v is the initial total velocity and theta is the angle the particle is fired from.

At 45 degrees, sin (theta) = cos (theta) = sqrt(2) / 2
Therefore if its initial velocity was 141 m/s at 45 degrees, its velocity at the top of the curve would be 141 m/s * sqrt(2) / 2 = 99.7 m/s.

Answer 2

Heres how you approach this.

Since this is a vertical flight, the "top of the trajectory" is the point at which the bullet projectile starts to fall back. Here the speed is zero

For 45 degree.

1. Calculate the speed of the bullet in both the direction . Ie. along the x-axis ( horizontal) and Y-axis ( vertical ).

At the top of the flight , speed along y-axis is zero. Speed along x-axis is 142 * sqrt(1/2) which is 100 m/s rounded

Answer 3

at top of its trajectory only horizontal component of velocity will present so
velocity at top of trajectory=u cos45
=100m\s

Answer 4

Go to your physics text book and look up the appropriate formula and plug in those numbers.