Physics, asked by zehramadani1593, 23 days ago

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

Answered by anjalirehan04
3

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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Answered by DakshRaj1234
1

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

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