An artillery shell is fired from a cliff 20.7 m above a valley with a velocity of 255 m/s at an angle of 37.7 degrees above the horizontal

Question

How far (in km) has the shell traveled horizontally when it is at its maximum height?

Answer 1

Its equation for the height in meters (time t in seconds) is

h(t) = -4,9t² + 255sin(37.7°)t + 20.7

Maximum height is at the vertex of the height-vs.-time parabola, which occurs at

t = -b/(2a)

Plug in

b = 255sin(37.7°)

a = -4.9

and get the time at maximum height. Then multiply this time by the constant horizontal speed, which is

v_x = 255cos(37.7°)

That product will give the horizontal travel distance in meters at maximum height. Plug in the numbers, convert the result to km, and finish from here.

Answer 2

To determine how far the shell has traveled horizontally, we need to know ho long it took to get there. Whenever we're talking about projectile motion, the velocity in y is zero at the maximum height.

Recall that v_y = v*sin(θ) and v_x = v*cos(θ)

If the initial velocity in y (v_y) it 225m/s * sin (37.7 deg), the velocity at the maximum height is 0, and gravity is 9.81 m/s², we can use v_f=v_i+a*t to get the time.

0=225m/s * sin (37.7 deg) + (-9.81 m/s²)*t

Solve for t. Once that's done, we can use d=v_x*t to get the distance we are asked for.