A pistol that fires a signal flare gives it an initial velocity (muzzle velocity

Question

A pistol that fires a signal flare gives it an initial velocity (muzzle velocity) of 194 m/s at an angle of 50.0 above the horizontal. You can ignore air resistance.

1-Find the flare's maximum height if it is fired on the level salt flats of Utah.

2-Find the distance from its firing point to its landing point if it is fired on the level salt flats of Utah.

3-Find the flare's maximum height if it is fired over the flat Sea of Tranquility on the Moon, where g=1.62m/s2.

4-Find the distance from its firing point to its landing point if it is fired over the flat Sea of Tranquility on the Moon, where g=1.62m/s2.

Explanation / Answer

As the pistol is shot, the trajectory is in the shape of a parabola.

First we need to define the initial velocity (v_0) in the horizontal and vertical directions. Since we know the angle of shooting, we can find these velocities using trig. functions.

v_0_x = 194*cos(50) = 124.11 m/s
v_0_y = 194*sin(50) = 148.54 m/s

Since the trajectory is in the shape of a parabola, we need to find the point where the projecticle reaches the max (i.e. velocity = 0)

v_f_y = (a*t) + v_0_y

Using v_f_y = 0 and a = 9.8, we get t = 15.46 sec.

Now we can find the horizontal displacement when the projectile reaches its peak:

x = [v_0_x]*t+0.5*a*t^2

Since we are calculating the horizontal components, a = 0 (gravity only moves in vertical direction)

Therefore x = (124.11)(15.46) = 1917.5 m

Since the projecticle motion is parabolic in nature, it will take the same amount of time for the projecticle to hit the ground.

Therefore the total horizontal displacement = 2*1917 = 3834 m

For Sea of Tranquility, just plug in a=1.62 instead of 9.8 in the above equations.

When i do that, i get t= 94.71 s and x = 10050 m

Therefore, horizontal range = 10050*2 = 20100 m

As the pistol is shot, the trajectory is in the shape of a parabola.

First we need to define the initial velocity (v_0) in the horizontal and vertical directions. Since we know the angle of shooting, we can find these velocities using trig. functions.

v_0_x = 194*cos(50) = 124.11 m/s
v_0_y = 194*sin(50) = 148.54 m/s

Since the trajectory is in the shape of a parabola, we need to find the point where the projecticle reaches the max (i.e. velocity = 0)

v_f_y = (a*t) + v_0_y

Using v_f_y = 0 and a = 9.8, we get t = 15.46 sec.

Now we can find the horizontal displacement when the projectile reaches its peak:

x = [v_0_x]*t+0.5*a*t^2

Since we are calculating the horizontal components, a = 0 (gravity only moves in vertical direction)

Therefore x = (124.11)(15.46) = 1917.5 m

Since the projecticle motion is parabolic in nature, it will take the same amount of time for the projecticle to hit the ground.

Therefore the total horizontal displacement = 2*1917 = 3834 m

For Sea of Tranquility, just plug in a=1.62 instead of 9.8 in the above equations.

When i do that, i get t= 94.71 s and x = 10050 m

Therefore, horizontal range = 10050*2 = 20100 m

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