A rocket is fire at ground level targets 600m away with an initial velocity 85 metre per second find the two possible value of launch angle calculate the minimum time to hit the target
Answers
R= (85×85)sin2theta /g
600= (7225)sin2theta/9.8
600×9.8 = (7225) sin2theta
5880/7225 = sin2theta
0.81 = sin2theta
sin inverse (0.81) = 2theta
54 = 2theta
54/2 = theta
27° = theta
now this is the minimum value of angle theta to find the maximum value
theta max = 90° - 27°
theta max = 63°
now to find the minimum time we have to take the minimum value of theta
T minimum = 2(85) sin(27)/9.8
T minimum= (170)(0.45)/9.8
T minimum = 7.9seconds
so the minimum time to hit the target is 7.9seconds
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Answer:
R= (85x85)sin2theta /g
600= (7225)sin2theta/9.8
600×9.8 (7225) sin2theta
5880/7225 = sin2theta
0.81= sin2theta
sin inverse (0.81) = 2theta
54 = 2theta
54/2 = theta
27° = theta
now this is the minimum value of angle theta to find the maximum value
theta max = 90° -27°
theta max = 63°
now to find the minimum time we have to take the minimum value of theta
T minimum = 2(85) sin(27)/9.8
T minimum= (170)(0.45)/9.8
T minimum = 7.9seconds
so the minimum time to hit the target is 7.9seconds
please mark me brain mar list