First vertical problem:
Vi = 75.2 sin 34.5 = 42.6 m/s
v = Vi - g t
g = 9.81 m/s^2
so
v = 42.6 - 9.81 t
at the top, v = 0
t = 42.6/9.81 = 4.34 seconds to top
2 t = 8.68 seconds total in air
range = 8.68 * 75.2 cos 34.5
at t = 1.5
v = Vi - g t = 42.6-9.81(1.50)
= 27.9 m/s
u = 75.2 cos 34.5 = 62 m/s
speed = sqrt (u^2+v^2
= sqrt (62^2 + 27.9^2)
tan angle = v/u = 27.9/62
angle = 24.2 degrees above horizontal
I'm having trouble where to start with this question.. Any help would be helpful.
A projectile is fired with an initial speed of 75.2 m/s at an angle of 34.5° above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered (that is, the range), and (d) the velocity of the projectile 1.50 s after firing.
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