What is the maximum height of the projectile if it is launched at 114 m/s and at an angle 45 degrees to the horizontal.
The initial velocity V at which a projectile making an angle of 45 degrees with the horizontal is launched can be divided into the horizontal component equal to V*cos 45 and the vertical component equal to V*sin 45. Assuming the absence of any frictional forces, the time during which the projectile is traveling upwards is equal to the time during which it is traveling downwards.
The projectile launched at 114 m/s at an angle of 45 degrees to the horizontal travels for a total duration t such that 0 = 114*sin 45 - 9.8*t
=> t = [114*(1/sqrt 2)]/9.8
=> t = 8.225 s
The time spent moving upwards is half of this or 4.11 s
In 4.11 s the vertical distance moved by the projectile is 114*(1/sqrt 2)*4.11 - 0.5*9.8*4.11^2 = 248.53 m
The maximum height of the projectile launched at 114 m/s at an angle of 45 degrees with the horizontal is 248.53 m.