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Question

1. What is the maximum height of a bullet shot from a gun at 150 m/sec at a 5.0 degree angle?

2. In order to measure the acceleration due to gravity on the moon, an astronaut fires a projectile from a launcher. The initial velocity is 57.0 ft/sec at 22.5 degrees and the projectile's range is measured to be 220.0 feet. What is the gravity on the moon as calculated by this data? Provide your answer in units of feet and seconds.

3. A submarine travels at 18.0 mph for 92.0 minutes at 90.0 degrees. Then the sub adjusts its course, traveling at 16.0 mph for 2.25 hours at 155 degrees. What is the sub's resultant vector? Units of miles and hours should be used to solve this problem.

4. What is the range of a gun that fires its bullets with a velocity of 175 meters per second at an angle of 30.0 degrees? Assume the bullets land at the same height from which they were launched. Use units of meters and seconds to solve this problem.

5. Aimed at an angle of 45 degrees, a cannon has a range of 4,123 feet. What is the initial speed of a cannonball launched from this cannon? Assume the cannonballs land at the height from which they were launched. Use units of feet and seconds to solve this problem.

6. A stuntman was prepared to jump a river in a car. From the base of a 33.0-foot cliff to the other side of the river was 113 feet. How fast should his initial velocity be in order to clear the river and what is the y-component of his velocity right before he hits the ground on the other side of the river? The ground leading up to the cliff edge is level and no ramp is being used. Use units of feet and seconds to solve this problem.

7. A hiker follows a trail map with the following instructions on how to get to a waterfall: Travel with a constant velocity of 2.1 meters per second at a heading of 120.0 degrees for 1.1 hours, then turn to a heading of 200.0 degrees and travel at the same speed for 2.3 hours. What will the hiker's final displacement be? Calculate your answers using units of meters and hours.

Answer 1

Solution:

1- Given that the initial velocity is V₀ = 150 m/s, then the vertical component of the initial velocity is

V_oy = 150 sin ∝ = 150 sin(5)= 13.07 m/s

From the equation V_fy² - V_oy² = -2gy knowing that the final velocity at max height is 0 , V_fy = 0

the equation reduces to -V_oy² = -2gy , and y = V_oy² / 2g = (V_osin∝)² / 2g = V_o² sin ²∝ / 2g

y = (13.07 m/s)² / 2(9.8 m/s²) = 8.72 m

2- The equation for the range R = V_o² sin 2∝ / g , solving for g = V_o² sin 2∝ / R

g = ( 57 ft/s)² sin( 2(22.5)) / 220 ft = 10.44 ft/s²

3- The distance the boat travels at 90° is 18 mi/hr ( 92 min / 60 min/hr) = 27.6 mi

The distance that the boat travels at 155° is 16 mi/hr ( 2.25 hr) = 36 mi

Adding the two vectors graphically creates a resultant R that can be computed using the cosine law

The angle between the two vectors is graphically determined as 360°°- 155°- 90° = 115°

R = √[(24.6 mi)² + ( 36 mi)² - 2(24.6 mi)(36 mi) cos (115°)] = 53.8 mi

4- This is similar to problem 2 and you need to find the range R = V_o² sin 2∝ / g

R = (175 m/s)² sin( 2*30) / 9.8 m/s² = 2706 m

Other method of solving this is the horizontal component of the force vector times the time is also the range

such that R = X = V₀ cos ∝ t = (175 m/s) cos (30)(t) = (151.5 m/s) t and you need t to solve for the range

Recall that if the beginning altitude and the final altitude is the same then y = y_o

In the general equation y = -1/2 g t² + V₀sin ∝ t + y₀ you can eliminate y and y_o from both sides

so rearranging the equation 0 = -1/2 gt² + V_osin ∝ t, 1/2 g t² = V_o sin ∝ t , g t ² = 2 V₀ sin ∝ t , and

g t = 2 V_o sin ∝, t = 2 V₀ sin ∝ / g = 2 ( 175 m/s)(sin 30)/ 9.8 m/s² = 17.86 s

and R = (151.5 m/s)(17.86 s) = 2705.8 m

Compared with method 1 we had R = 2706 m Close enough

5- This problem is similar to the one above except that you need to find the initial velocity V_o

The range R = V_o² sin 2∝ / g rearrange to solve for V_o , g R = V_o² sin 2∝ , V_o² = g R / sin 2∝

Remember the g in US units is 32.2 ft/s² V₀² = (32.2 ft/s²)(4123 ft) / sin (2*45) = 132,761 ft²/ s²

V₀ = √ ( 132,761 ft²/s²) = 364.4 ft/s

another method is recognizing that the Range is the horizontal component of the velocity vector times the time R = V_o cos ∝ t , V_o = R / cos ∝ t ----- Eqn 1 , so you need t to solve

the other equation is 0 = -1/2 gt² + V_o sin ∝ t , and solving for V₀ = g t / 2 sin ∝ ------ Eqn 2

placing Eqn 1 equal to Eqn 2 ; R / cos ∝ t = g t / 2 sin ∝ , cos ∝ g t² = 2 sin ∝ R, and

g t² = 2 sin ∝ R / cos ∝ = 2( sin ∝/ cos∝ ) R = 2 tan ∝ R ; t² = 2 tan ∝ R / g

t² = 2 ( tan 45) ( 4123 ft) / 32.2 ft / s² = 256 s², t = √256 s² = 16 s and

V₀ = ( 4123 ft) / (cos (45 )) 16 s = 364.4 ft/s same answer as in the first method.

6- in this problem you know the range R = 113 ft, and the initial altitude y_o = 33 ft the height of the cliff and you are asked to find V_o at takeoff from the cliff. The car did not use a ramp, and this means that the angle of take off was 0 wrt to the horizontal at the cliff.

The Range R = V₀ cos ∝ t = V₀ cos 0 t = V₀ t

and y = -1/2 g t² + V_o sin ∝ t + y_o knowing that the final altitude is y =0 at the ground

0 = -1/2 g t² + V_o sin (0) ( t) + 33 ft ; 0 = -1/2 g t² + 33 ft Rearranging , 1/2 g t ² = 33 ft

solving for t, g t² = 66 ft, t² = 66 ft / g = 66 ft / 32.2 ft /s² = 2.05 s², and t = √2.05 s² = 1.43 s

R = V₀ t, V₀ = R / t = 113 ft / 1.43 s = 79.0 ft/s

The final velocity can be found from the equation just before the car hits the ground

Vf² - V_o² = 2 g y_o , Substituting V_f² - (79.0 ft/s)² = 2 ( 32.2 ft/s²)(33 ft) = 2125.2 ft²/ s²

and V_f² = 2125.2 ft²/s² + ( 79.0 ft/s)² = 8366.2 ft²/ s² , and V_f = √8366.2 ft²/ s² = 91.5 ft/s c