3. (The algebra for this problem is... challenging. But it is just algebra. Help

Question

3. (The algebra for this problem is... challenging. But it is just algebra. Help will be posted to the website soon.) After a heavy snowfall, a ski resort uses explosives to intentionally create a controlled avalanche. (They usually do this at night... although in March 2016 in Mammoth they were doing it in the morning while we were on the slopes!) The location from where an explosive shell is fired is at an elevation of 2200 m and a direct distance of 2100 m from the target snow, which is at an elevation of 3000 m. If the shell is launched with an initial speed of 212 m/s find:

(a) the time it is in the air;

(b) the angle at which it was launched.

Explanation / Answer

As per provided info: target snow is the peak point of this projectile motion and at the peak point vertical velocity of shell will be ZERO. As per given data we can understand situation that canon is at 2200m elevation and target is at 3000m elevation so target is at (3000-2200)=800m of vertical height from canon. And target is at 2100m of straight line distance from canon. So here we can take it as a right angle triangle in which hypotenuse is 2100m and perpendicular is 800m then base can be found out. The base will be horizontal distance between canon and target. Value of base is 1941.64m using Pythagoras formula.

Now formula for vertical velocity of a projectile is: Vy= Vy0 -gt,

Where Vy is vertical velocity at any point of time, Vy0 is initial vertical velocity and g=acceleration due to gravity and t is time.

Suppose shell was fired making angle x° with horizontal.

So we take Vy= 0 when shell reaches target.

Now initial vertical component of velocity=Vsinx°=212sinx°

Putting these values we have: 0=212sinx°-9.8*t

So, t=212sinx°/9.8. ............ equation (1)

Using Newton's formula V=u+ a*t, . We have:

0=212sinx°- 9.8*t.... equation (2). Gravitational acceleration is negative since it's acting opposite to motion of projectile.

Now time t in eauation(1) is same for horizontal flight time of shell. So using formula time * velocity= displacement. we have:

Horizontal velocity * t = horizontal distance between canon and target

i.e 212cosx° * (212sinx°/9.8) = 1941.64

Sinx°*cosx°=(1941.64* 9.8)/212^2

(Sin2x°)/2=0.4233

x°=28.92°

Putting this value of x°=28.92 In equation (1) we get:

t=10.45secs.

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