The vertical component is vy = 40-9.8t
so the angle θ at t=7 is tanθ = (40-9.8*7)/30 = -28.6/40
θ = 144.44°
A projectile is fired on level ground, and has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively. After 7 seconds of flight, what is the approximate angle of the projectile's velocity vector with respect to horizontal?
So I figured out the angle to the horizontal the projectile is fired at is 53 degrees, but I don't know much else to do :(
2 answers
Vo = 30+40i = 50[53o]
yo+g*T = 0
40-9.8T = 0
T = 4.1 s. to reach max ht.
Y = Yo+gT = 0+9.8*(7 -4.1) = 28.4 m/s.
Tan A = Y/X = 28.4/30
A = 43.5o
yo+g*T = 0
40-9.8T = 0
T = 4.1 s. to reach max ht.
Y = Yo+gT = 0+9.8*(7 -4.1) = 28.4 m/s.
Tan A = Y/X = 28.4/30
A = 43.5o
1 answer