A projectile is fired from the origin (at y = 0 m) as shown in the diagram.

Question

A projectile is fired from the origin (at y = 0 m) as shown in the diagram. The initial velocity components are V0x=310m/s and V0y=26m/s The projectile reaches maximum height at point P, then it falls and strikes the ground at point Q, which is 20 m below the launch point. What is the horizontal distance that the projectile travels (labeled x in the diagram)?

700 m

3.2 km

870 m

1.3 km

1.9 km

Answer 1

Hey Johnny!

In this problem, we are dealing with kinematics - the movement of an object. This object is shot diagonally so it has both a horizontal velocity, v0x, and a vertical velocity, v0y. Let's draw a picture:

→ v0x (unchanged, due to no horizontal forces)

· Point P (peak height)

/ \

/ \

/ \

↑v0y \

· → v0x \ *point of interest

(x = 0, y = 0, t = 0) |

· Point Q (x, y = -20, t)

*At the point of interest, the object is at falling at the same height at which it was launched. The physics works out that at this point it is falling, rather than rising, but with the same magnitude of v0y. Gravity acted against it for an amount of time, then for an equal amount of time gravity brought it back to the same speed.

The goal is the horizontal distance, and we are given the horizontal velocity, v0x = 310 m/s. So, what we need to do is find the total time, t, that the object is in the air. We can do so using the vertical velocity, gravity, and the final height. Let's set up the equations:

1. y = ½*a*t² + v0y*t + y0
2. x = ½*0*t² + v0x*t + x0
3. x = v0x*t + x0

Equation 1 tells us the height the object will be over time, but ignores the horizontal distance it has traveled. Equation 2 deals with the same physics, but has 0 acceleration, so it simplifies to a velocity-only equation. We can use Equation 1 to solve for time, t, and then plug that into Equation 2.

First we'll plug in what we know into Equation 1 and solve it for t:

y = -20 m (Point Q)

a = g = -9.8 m/s² (negative since it accelerates downward)

v0y = 26 m/s

y0 = 0 m (origin)

-20 m = ½ * -9.8 m/s² * t² + 26 m/s * t + 0 ; Simplify and move the 20 m over

0 = -4.9 m/s² * t² + 26 m/s * t + 20 m

That looks hard to factor, so let's use the quadratic formula to solve for roots.

a = -4.9

b = 26

c = 20

t = -b ± √ (b² - 4ac) = -26 ± √ (26² - 4*-4.9*20)

2a 2 * -4.9

= -26 ± √ (676 + 392)

-9.8

= -26 ± 32.68

-9.8

t = -0.68 s, 5.99 s

A negative answer for time in the air makes no sense at all, so we find that t ≈ 6 s in the air total. We can now plug that into Equation 2 and solve for x:

v0x = 310 m/s

t = 5.99 s

x0 = 0 (origin)

x = v0x*t + x0

= 310 m/s * 5.99 s + 0

= 1856.2 m

≈ 1.9 km

The object lands almost 2 km away, which makes sense due to the high velocities! Be sure to follow your instructor's rules for sig figs.

Hope this helps! Please reach out with any further questions.

Answer 2

x = 310t; y = 26t - 4.9t². To get time t of flight let set y = - 20.

We get equation for time 26t - 4.9t² = - 20; 4.9t² - 26t - 20 = 0. From here t = 6 s

and horizontal distance x = 310 m/s·6s = 1860 m ≈ 1.9 km