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[SOLVED] projectile shot downward
A certain airplane has a speed of 255.0 km/h and is diving at an angle of 30.0° below the horizontal when a radar decoy is released (see Figure 4-36). The horizontal distance between the release point and the point where the decoy strikes the ground is 700 m. Neglect air resistance.
so Velocity in meters per second is 255*.2777=70.83m/s
Velocity in X direction is 70.83sin60=61.34m/s
Figure 4-36
(a) How high was the plane when the decoy was released?
t=m/v = 700/61.34 = 11.41s
Y=Yo+VoT+.5(a)(t^2)
Y=(0)+(0)(t)-4.9(11.41^2)
Y=638.1m which is wrong, can anyone help me out
A certain airplane has a speed of 255.0 km/h and is diving at an angle of 30.0° below the horizontal when a radar decoy is released (see Figure 4-36). The horizontal distance between the release point and the point where the decoy strikes the ground is 700 m. Neglect air resistance.
so Velocity in meters per second is 255*.2777=70.83m/s
Velocity in X direction is 70.83sin60=61.34m/s
Figure 4-36
(a) How high was the plane when the decoy was released?
t=m/v = 700/61.34 = 11.41s
Y=Yo+VoT+.5(a)(t^2)
Y=(0)+(0)(t)-4.9(11.41^2)
Y=638.1m which is wrong, can anyone help me out
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