# How long does it take for the projectile to get to its maximum height?In terms of initial velocity and initial angle, what is the time it takes for a projectile to get to its maximum height?...

How long does it take for the projectile to get to its maximum height?

In terms of initial velocity and initial angle, what is the time it takes for a projectile to get to its maximum height?  Assume acceleration is  -g.

mwmovr40 | College Teacher | (Level 1) Associate Educator

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To do this without Calculus we use the two projectile motion equation:

R = VxCos(theta)T

H = VxSin(theta)T - 1/2 gT^2

Where R is maximum range, T is maximum time of flight, V is initial velocity  and theta is the angle of launch.

Solving the first equation for T gives:

T = R/VCos(theta)

at R, H = 0, so we can plug T into second equation and solve for V:

0 = VSin(theta)R/Vcos(theta) - 1/2g (R/VCos(theta))^2

0 = RTan(theta) -1/2 g(R^2/V^2Cos^2(theta))

RTan(theta) = 1/2 g (R^2/V^2Cos^2(theta))

Bringing all but V^2 to the same side gives

1/V^2 = 2RTan(theta)Cos^2(theta)/(gR^2) = 2Sin(theta)Cos(theta)/(Rg)

You can now take the reciprocal of both sides of the equality and then the square root of both sides to solve for V.

With V, you can substitute into the equation for T to get the maximum time of flight.  To get the time of flight for the maximum height, divide the maximum time in half.

V = sqareroot (Rg/2sin(theta)cos(theta))