Projectile Fired at an Angle - Problems II (AP) - Projectile Motion

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AP Problem Set - combining vertical and horizontal kinematics to solve angled-launch projectiles, including quadratic time-of-flight solutions

Warm-Up

The football has 2 velocities. Which velocity is the football player matching?

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Horizontal Velocity

The football and the football player have a constant horizontal velocity

Show Me The Physics

Problem 1

Given

A projectile leaves the ground at an angle of 60.° and a speed of 100. m/s (level ground).
Find: (a) the object's maximum height & (b) total time in the air.

(a) First find the initial vertical component of the object's velocity

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V_i = 100. m/s, θ = 60.°, V_iy = ?

V_iy = V_i(sinθ)
= (100. m/s)sin60.°

V_iy = 87. m/s up

(a) Now find the maximum height

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V_iy = 87. m/s, a_y = -10. m/s² (ascending), V_fy = 0 (at max height), d_y(max) = ?

V_f² = V_iy² + 2ad →
0² = (87. m/s)² + 2(-10. m/s²)d

0 = 7569 - 20.d_y →
d_y = 378 m

d_y = 380 m (2 sig figs)

(b) Total time in the air

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V_iy = 87. m/s up →
t = 2(87 m/s) / 9.8 m/s²

t = 18 sec

Problem 2

Rock thrown from a cliff at 45 degrees below the horizontal

Given

A rock is thrown from a cliff with an initial speed of 40. m/s at an angle of 45.° below the horizontal.
Find: (a) if the rock strikes the ground in 1.0 sec, the height of the cliff.

(a) What is the vertical component of the initial velocity?

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θ = 45.°, V_i = 40. m/s, V_iy = V(sinθ) = (40. m/s)sin45.°

V_iy = 28 m/s down

(b) If the rock strikes the ground in 1.0 sec, what is the height of the cliff?

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d_y = ? (height), a_y = +10. m/s², t = 1.0 sec, V_iy = 28 m/s (from before)

d_y = V_iyt + (½)at²
= 28. m/s(1 sec) + ½(10. m/s²)(1.0 sec)² = 28 m + 5.0 m

d_y = 33 m down

Problem 3

Ball launched at 45 degrees with equal horizontal and vertical velocity components of 9.0 m/s, elapsed time 0.92 s at peak

Given

A ball is launched at 45.° with V_ix = V_iy = 9.0 m/s.

(a) At peak height, V_x = ?

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V_x = 9.0 m/s

(b) Total time in the air?

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1.84 sec (twice the time to peak)

t = 1.8 sec (2 sig figs)

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9 m/s, V_x is constant

Problem 4

Given

A wild bowler releases his bowling ball at a speed of 15 m/s and an angle of 34.0° above the horizontal, leaving his hand at a height of 2.20 m above the ground.
Find: the horizontal distance the bowling ball travels.

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d_x = V_xt
= Vcosθt — need to find t

V_iy = Vsinθ
= 8.6 m/s

d_y = V_iyt + (½)at² →
2.20 m = (-8.6 m/s)t + 4.9t²

0 = 4.9t² + (-8.6 m/s)t - 2.20 m — Quadratic Equation

t = 1.99 s, -0.225 s (reject the negative root)

d_x = Vcosθt
= (15 m/s)cos34(1.99 sec)

d_x = 25.6 m

Problem 5

Given

The ammunition fired from an M-16 rifle has a muzzle velocity of 1000 m/s. A sniper perched on a tower 50 m high aims at a target and fires. Neglect air resistance.

(A) The sniper aims her rifle 53.1° above the horizontal. How far away from the platform does the bullet hit the ground?

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V_x = Vcosθ = 600 m/s →
d_x = Vcosθt = [600 m/s]t — need to find t

Make down positive: Δd_y = -Vsinθt + ½gt²

50 m = -1000 m/s[sin53.1]t + 4.9t²
= -800t + 4.9t²

0 = 4.9t² - 800t - 50. m →
t = 163 sec

d_x = Vcosθt
= [600 m/s]t

d_x = 98000 m

(B) With what speed would the bullet hit?

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V_x = 600 m/s (down negative)

V_fy = -Vsinθ + gt
= -800 m/s + 10 m/s²(163 sec) = 830 m/s

Pythagorean Theorem

V = 1000 m/s

(C) If the target is 500 meters away on a platform 50 meters high, what angle would the sniper have to make with her gun to hit the target?

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D_x = V_xt →
500 m = [(1000 m/s)cosθ]t

Δd_y = V_iyt + (½)at² →
0 = Vsinθt + 5t²

θ = 44.9°

Summary

(A) 98000 m · (B) 1000 m/s · (C) 44.9°

The Monkey and the Hunter

Monkey and Hunter — Don Ion

(offsite)

Flash Physics — Don Ion

St. Mary's U. Astronomy and Physics Dept.

| Onsite |

Monkey and Hunter (TeacherTube video)

Projectile Motion Game — St. Mary's Physics

Continuing your studies? Intro to Circular Motion

Check Your Understanding

5 quick questions — pick an answer to see if you’re right.

1. A projectile is launched at 100. m/s at 30.° above level ground. Using g = 10. m/s², what is its total time in the air?

2. For that same projectile (Viy = 50. m/s, g = 10. m/s²), what is its maximum height?

3. A rock is thrown from a cliff with Viy = 20. m/s downward and hits the ground 1.0 s later. Using g = 10. m/s², how high is the cliff?

4. A ball is launched at 45° with Vix = Viy = 9.0 m/s. What is its speed at the peak of its path?

5. Solving a launch-from-a-height problem with the quadratic formula gives t = 1.99 s and t = −0.225 s. What is the time of flight?

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