Question

A cannon placed on a wall 64 feet above the ground fires a cannon ball level in the horiznotal direction with horizontal velocity 80. How far (the horizontal distance) from the foot of the wall does the cannon ball land when it hits the ground?

Answer 1

Answer:

160 ft

Explanation:

As the canon ball was shot horizontally, its initial vertical velocity is 0. The ball vertical motion is generated by the gravitational acceleration g = 32m/s2. We can calculate the time it takes for it to drop 64 ft

$s_v = gt^2/2$

$t^2 = 2s/g = 2*64/32 = 4$

$t = \sqrt{4} = 2s$

This is also the time it takes to travel horizontally at a constant rate of 80 ft/s if we ignore air resistance.

$s_h = vt = 80 * 2 = 160 ft$

So the ball would land 160ft away from the wall on the ground

Answer 2

Answer:

159.57 Feet Horizontally from the foot of the wall

Explanation:

Given Data:

Height = $h$ = 64 feet

gravitational acceleration = g = 32.17 $ft/s^{2}$

Initial Vertical Velocity component $Vi = 0$

Initial Horizontal Velocity Component $Ui$ = 80 $ft./s$

To find = Time taken to reach ground = $t$

Horizontal Distance from the foot of the wall = $x$ = ?

Calculation:

From the equation of motion $h = Vi*t + 1/2*g*t^{2}$ ........ (1) (in terms of Vertical component of velocity).

Putting in values in the above equation.

$t = 1.99 s$ (Time taken to reach the ground)

Now, to find $x$

we use the same  equation of motion we used above but this time we solve for the horizontal component of Velocity

$x = Ui*t + 1/2*a*t^{2}$

$a = 0\\Ui = 80 ft/s\\t = 1.99 s$

Putting in values in the above equation.

$x = 80*1.99\\x = 159.57 ft$