# That darn PHYSICS again

posted by
**Sean** on
.

Hi!

i need some help with this ASAP!

a body is projected upwards from a point on level ground with a velocity of 45 m/s at an elevation of 30 degrees. Working from the expression for uniformly accelerated motion in a straight line, determine

1) the time of flight

2) the distance from the point of projection to the point where the body strikes the ground

3) the greatest height reached

THANKS VERY MUCH!@

The vertical component of initial velocity is

Vyo = 45 m/s * sin 30 = 22.5 m/s

The time required to hit maximum height is t1 = yo/g and the time of flight is twice that.

The horizontal coordinate where the body hits the ground is

X = V cos 30 * (time of flight),

where V = 45 m/s

The maximum height is acquired when

Vyo = 0, at t = t1. The vertical height at that time is

V sin 30 * t1 - (g/2) t1^2

Hi!

i need some help with this ASAP!

a body is projected upwards from a point on level ground with a velocity of 45 m/s at an elevation of 30 degrees. Working from the expression for uniformly accelerated motion in a straight line, determine

1) the time of flight

2) the distance from the point of projection to the point where the body strikes the ground

3) the greatest height reached

THANKS VERY MUCH!@