Question

Calculate the time-of-flight for a projectile launched at an angle to the horizontal if the velocity is 141.4m/s and the angle is 45 degrees

Answer 1

@amistre64

Answer 2

The time of flight is when y position of object is equal to zero, therefor using the equation of motion: \[y=-\frac{ 1 }{ 2 }gt ^{2}+V _{o}y(t) \]
and the given info:

V = 141.4 m/s
angle = 45

first we calculate the Y component of velocity:
Vy = 141.4 * Sin(45) = 99.98 m/s

then we substitute the Y component of velocity and y position (which is zero) and "g" (g = 9.8 m/s) in the above equation to get the following:
\[0=-\frac{ 1 }{ 2 }(9.8)t ^{2}+99.98t \]

not its just algebra, first divide the whole thing by "t", then take 99.98 to the other part, divide be "t" coefficient to get t

the answer is: t = 20.4s

Answer 3

Thank you :D