A projectile is fired with an initial speed of 65.2 m/s at an anmgle of 34.4 degrees above the horizontal on a long flat firing range.
Determine the max heigh reached by the prkectile, the total time in the air and the total horizontal distance covered and the velocity of the projectile 1.50 seconds after firing.
The Attempt at a Solution
I am going to check the first parts now, and then continue once I am sure I am going in the right direction.
Vxo = vocos 34.5 = 65.2/0.824 = 79.1 m/s
Vyo = vosin34.5 = 65.2/0.566=115 m/s
y=Vyot - 1/2gt^2= 675 meters maximum height.
y=yo + vyot - 1/2 gt^2
0 = 0 + 115t - 1/2 x 9.80 x t^2
t = 0 at initial point
t= 2 (115)/9.80 = 23.5 seonds.