A projectile is fired with an initial speed of 46.6 m/s at an angle of 42.2 ∘ above the horizontal on a long flat firing range.
1-Determine the maximum height reached by the projectile.
2-Determine the total time in the air.
3-Determine the total horizontal distance covered (that is, the range).
4-Determine the speed of the projectile 1.50 s after firing.
5-Determine the direction of the motion of the projectile 1.50 s after firing.
v = vo +at x = xo + vot + .5at2 v2 = vo2 + 2a(x - xo) x=vt
The Attempt at a Solution
A.) Determine the maximum height Ymax reached by the projectile.
Ymax = (V*sin 45.2)^2/2g = (43.6*0.71)^2/19.612 = 48.86 m
B.)Determine the total time t in the air.
t = 2*V*sin 45.2/g = 2*43.6*0.71/9.806 = 6.314 sec
C.)Determine the total horizontal distance covered (that is, the range).
Δx = V*cos 45.2*t = 43.6*0.705*6.314 = 194.08 m
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