Question

A projectile is thrown with speed 40m/s at angle theta from horizontal

Answer 1

So,

If the projectile is thrown at an angle Ф with a velocity of 40 m/s,

I'm assuming we have to find the range , maximum height and time taken during the flight for the projectile ,

Now,

Ux and Uy are the velocity components in x and y direction

H=Uy²/2g

H= (u² sin²Ф )/ 2g

H=(1600 × sin²Ф)/20

H= 800 sin²Ф

For the info, Height will be maximum when angle Ф = 45°

R= (2 × Ux × Uy)/ g

R= ( u² × sin2Ф) /g

R = (1600 × sin2Ф)/10

R=160 sin2Ф

Time of flight

T= (2 × Uy)/ g

T= (2 × u sinФ)/ g

T= (2 × 40 × sinФ) /10

T= 8 sinФ

These are the required parameters