Physics, asked by zynubkhan97, 15 days ago

an artillery with maximum range 10-20 is to be fired for a target at 600 metre what should be the angle for maximum and minimum time what will be the time and of light in both cases what will be the maximum height in both cases?​

Answers

Answered by KISHANBHARATH
0

Answer:

sorry I don't know this answer

Answered by gyaneshwarsingh882
0

Answer:

Explanation:

Height of the fighter plane =1.5km=1500 m

Speed of the fighter plane, v=720km/h=200 m/s

Let be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.  

Muzzle velocity of the gun, u=600 m/s  

Time taken by the shell to hit the plane =t

Horizontal distance travelled by the shell =u  

x

t

Distance travelled by the plane =vt

The shell hits the plane. Hence, these two distances must be equal.

u  

x

t=vt

u Sin θ=v

Sin θ=v/u

   =200/600=1/3=0.33

θ=Sin  

−1

(0.33)=19.50

In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H) higher than the maximum height achieved by the shell for any angle of launch.

H  max

=u  2  sin  2

(90−θ)/2g=600 2 /(2×10)=16km

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