Question

two possible angles to hit a target by a motor shell fired with an initial velocity of 98 metre per second are 15 degree and 75 degree calculate range of projectile and minimum time required to hit a target​

Answer 1

Answer:

4.72 secondsGiven the distance of 490m, equal elevation, and starting velocity of 490m/s, there should be 2 solutions relatively close to horizontal and vertical launches. The vertical launch would take slightly less than 20 seconds from launch to impact. The horizontal launch would take slightly more than 5 seconds, so with “minimum time” in the question, this would be the angle you want.

Iteratively this should account for about 5 seconds of freefall distance which is appx 123m, so the launch angle would be close to 15 degrees above the horizontal axis.

To calculate this time with ballistic precision math stuffs, we have a variable angle of launch “Q”, which translates to an initial velocity divied into X and Y vectors.

98*Cos(Q)= X0ms

98*Sin(Q)= Y0ms

450X0=t

This is true because the horizontal X0 velocity remains constant, so however many seconds it takes for this to reach horizontal destination of 450m, that is the same time it must take for the Y0 vertical velocity to drop to its negative equivalent and hit the target.

Y0∗29.8=t

Applying gravity as 9.8 meters per second^2 it will take Y0/9.8 seconds to reach a velocity of zero at the top of the parabolic trajectory. Then it will take the same amount of time again to drop to the target, hence the *2

450X0=Y0∗29.8

After solving for t based on X and Y, set them equal to eachother. Now simplify.

450∗9.82=X0∗Y0

2205=982∗Cos(Q)∗Sin(Q)

45/196=Cos(Q)∗Sin(Q)

Trig properties n stuff

45/196=0.5∗Sin(2Q)

45/98=Sin(2Q)

Sin−1(4598)=2Q

Sin−1(4598)2=Q

Q=13.66 degrees

Now we know the launch angle for minimum time. Plug this in to the above equations to find t=?

t=4.72 seconds

Happy homeworking!