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# Projectile Motion Problem?

A shell leaves a mortar with a muzzle velocity of 500ft/s directed upward at 30degrees with the vertical. Determine the position of the shell and its resultant velocity 20s after firing. How high will it rise?

### 3 Answers

- electron1Lv 79 years agoFavorite Answer
Final vertical position = Initial vertical velocity * time – ½ * g * time^2

Final vertical velocity = Initial vertical velocity – g * time

http://en.wikipedia.org/wiki/Standard_gravity

According to the website above, the acceleration due to gravity = -32.174 ft/s^2

As the mortar moves upward, it decelerates at 32.174 m/s^2.

Initial vertical velocity = vi * cos θ = 500 * cos 30˚ = 433 ft/s

At 20 seconds, vertical velocity = 433 – 32.174 * 20 = -210.48 ft/s

Since the velocity is negative, the mortar has moved up to its highest position and is now moving downward at 210.5 ft/s.

At 20 seconds, the vertical position = 433 * 20 – ½ * 32.174 * 20^2 = 6434.8 ft

During the same time the horizontal velocity remains constant.

Horizontal velocity = v * cos θ = 500 * sin 30˚ = 250 ft/s

At 20 seconds, the horizontal position = 250 * 20 = 5000 ft

At 20 seconds, the magnitude of the velocity = (-210.48^2 + 250^2)^0.5 = 326.8 ft/s

The tangent of the angle from vertical = 250 ÷ 210.48

Angle from vertical = 49.9˚

At the maximum height, the vertical velocity is 0 ft/s

vf = vi – g * t

0 = 433 – 32.174 * t

433 = 32.174 * t

Time to reach maximum height = 13.46 seconds

Final vertical position = Initial vertical velocity * time – ½ * g * time^2

Maximum height = 433 * 13.46 – ½ * 32.174 * 13.46^2

Maximum height = 2913.7 ft

- AmyLv 79 years ago
Use trigonometry to figure out the vertical and horizontal components of the initial velocity.

Horizontal velocity remains constant.

Calculate the horizontal position after 20s using the equation x = vt

Vertical velocity is subject to a downward acceleration due to gravity.

Calculate the vertical position after 20s using the equation y = (v0)t - 16 t^2

(the -16 is the acceleration due to gravity in ft/s^2)

Also calculate the vertical component of velocity after 20s. Combine this with the horizontal component to find the speed and direction.

- oldschoolLv 79 years ago
Use Vf^2 = Vi^2 + 2*a*d where Vf = 0 at peak height, Vi = Vyi = 500sin60 = 433ft/s, a = -32ft/s^2 and d is the peak height

d = 433^2/64 = 2930ft

When does the mortar reach peak height?

Vy(t) = Vyi - g*t = 0 at peak height => t = 433/32 = 13.5s

Ignoring air resistance, the time to peak height is equal to the time from peak height back to launch height. Total flight time = 2*13.5 = 27s

Vy(20) = 433- 32*20 = -207ft/s which means downward.

Let height be Y(t)

Y(t) = Yi + Vyi*t - 4.9*t^2 where Yi=0, Vyi = 500sin60, t = 20s

Y(20) = 433*20 - 16*20^2 = 2260