Projectile Motion: Ball thrown upwards from position y above ground.

Obtain an expression for angle at maximum horizontal distance.

R=vcosQ/g(vsinQ + sqrt((vsinq)^2 + 2gy)

where R:range and y: position above ground.

Need to find expression for which value of Q maximum dR/dQ=0.

Question

Projectile Motion: Ball thrown upwards from position y above ground.

Obtain an expression for angle at maximum horizontal distance.

R=vcosQ/g(vsinQ + sqrt((vsinq)^2 + 2gy)

where R:range and y: position above ground.

Need to find expression for which value of Q maximum dR/dQ=0.

ketz · Answer

URGENTT

anonymous · Answer

welcome,  i try to solve it

anonymous · Answer

but theta about horizontal or vertical

ketz · Answer

The final answer is:

Qmax = sqrt((2gy + v^2)/(2gy + 2v^2))

ketz · Answer

I does not matter whether horizontal/vertical..

anonymous · Answer

sorry, it time for me to do homework for tomorrow, i will come back