Analyzing the Motion of a Projectile

Question

A projectile is fired from a cliff 941 feet above the water at inclination of 45 degrees to the horizontal, with a velocity of 610 feet per second. The h of the projectile can be modeled by

h(x) = (- 32 x^2) / 610^2 + x + 941

 

A) At what horizontal distance is the projectile at its maximum height?

B) What is the maximum height the projectile reaches?

C) At what horizontal distance from the firing point will the projectile hit the ground?

Answer 1

This problem can be approached using either a calculus or non-calculus approach.    The calculus approach is simpler, but I will use the non-calculus approach here.

 

The formula for h is quadratic in x (the horizontal distance), so the graph of h(x) is a parabola.   The standard formula for a parabola is   a x² + b x + c.    For this case  a =  -32/610² ,  b = 1 , c = 941.   The maximum height will be at the vertex of the parabola.   The x coordinate of the vertex of  a parabola in standard form is  x_v =  -b/(2a).  For this case this results in  x_v =  (1/2) 610²/32 =   5814.  

The maximum height is the y component of the vertex.   This can be obtained by substituting 5814 into the formula for h.   This results in y_v =  3848.

The projectile will hit the ground when h = 0.    This gives rise to the equation

0 = a x² + b x + c    with the values of a,b,c given above.   This equation is solved by the quadratic formula

 

x   =  ( -b - sqrt(b² - 4ac) /2a.      (I take the minus sign because the plus sign results in an unphysical negative value for x).      Plugging in the values of a, b, c (taking careful note that a is negative) gives   x =  12503