Question

A marter shell is fired at a ground level target 500 m
distance with an initial velocity of 90 m/s. What is its
launch angle?​

Answer 1

Answer:

20°

PLS MARK ME AS BRAINLIEST....

Answer 2

Answer:

You can just use the equations to get the result by using the formulas in other answers (or textbooks), but I think you get more understanding of physics by using (or imagining) an experimental approach to figure out what the angle is. The initial velocity (vector) of 90 m/s consists of an upward velocity and a horizontal velocity, which are independent of one another.

(Since you are looking for an angle, think of S=O/H C=A/H T=O/A.)

The upward velocity is 90 m/s times the sine of the angle.

The horizontal velocity is 90 m/s times the cosine of the angle.

First, let’s look at the horizontal travel only.

(The “decomposition” technique considers X and Y separately, then works on them one at a time. You can do this when X and Y are independent.)

Ignoring air friction, the only force involved is gravity, which does not affect horizontal travel. Therefore, we can calculate the time it takes for the shell to travel 500 meters: divide 500 m by 90 m/s to get time “t”:

500m90m/s=

509seconds=

5+59=5.5555…seconds

The time ( “t”) that it takes to travel 500 meters is the same as the time the shell takes to go up to its maximum altitude and fall back down (again ignoring air friction!), so let’s look next at what happens vertically

Assuming target and mortar are at (approximately) the same level, that means the shell rises to some maximum altitude, then falls down to the starting level after it has traveled 500 meters (in about 5.5 seconds).

Upward velocity of the shell starts at 90 m/s times sine of the angle.

After t/2 seconds, the upward velocity becomes zero, due to gravity.

Every second while the shell is climbing, the acceleration of gravity reduces upward velocity by approximately 10 m/s (until it becomes zero, at the maximum height).

The change in velocity (due to gravity) is (t/2) * 10 m/s

= 2.7777…seconds ∗10m/s

= 27.777…m/s

Therefore:

90 m/s times sine of the angle is equal to 27.777…m/s

sine of the angle is equal to 27.777…/90

sine of the angle is 0.30864…

the angle is approximately 0.31376 radians

Oh, you probably wanted the angle in degrees.

Also, the teacher probably wanted you to use 9.81 m/s2 instead of 10.

OK, just redo these steps with the constants you need.

(If you don’t get something near 18 degrees, check your math before handing it in.)