To start an avalanche on a mountain slope, an artillery shell is fired with an i

Question

To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 47.0° above the horizontal. It explodes on the mountainside 37.0 s after firing. What are the x and y coordinates of the shell where it explodes, relative to its firing point?

x =  m
y =  m

To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 310 m/s at 45.0° above the horizontal. It explodes on the mountainside 36.0 s after firing. What are the x and y coordinates of the shell where it explodes, relative to its firing point?
x = m
y = m

Explanation / Answer

v = 290 m/s ; theta = 47 ; t = 37 s

Let T be the time to reach the max height,

T = Vo sin(theta)/g

T = 290 x sin47/9.81 = 21.62 s

the shell travelled for the time, t' = T - t ; while going down

t' = 37 - 21.62 = 15.38 s

We know that, maximum height reached is given by:

H = Vo^2 x sin(2theta)/2g

H = 290^2 x sin(2x47)/2 x 9.81 = 4276 m

The vertical distance from the ground, while it travelled t' = 15.38 s

S = ut + 1/2 a t^2

S = 4276 - 0.5 x 9.81 x 15.38^2 = 3115.75 m

after 37 sec

D = 290 x cos47 x 37 = 7317.84 m

Hence, x = 7317.84 m and y = 3115.75 m

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