WORKSHEET 4: Projectile Motion

Problem 1: A rescue plane is flying at a constant elevation of 1200 m with a speed of 430 km/h (119 m/s) toward a point directly over a person struggling in the water. At what angle of sight f should the pilot release a rescue capsule if it is to strike (very close to) the person in the water?
Solution:
                    NOTE: The initial velocity of the capsule is the same as the velocity of the plane at the moment of release. Therefore v0 = 119 m/s in the horizontal direction. To solve this problem, we want to use trigonometry to solve for the angle. We have the height y = h = 1200m, so all we need is the x-distance.
                    To find the x-distance, we wish to solve: x = (v0 cos q)t. The only unknown in the equation is time so we need to find the time of flight.
                            y – y0 = (v0 sin q)t – ½gt2 is the other equation we can use, with y = -1200 m and q = 0
                            -1200 m = (119 m/s)(0)t – 4.90t2 ,
                            t2 = [(-1200 m)/(-4.90 m/s2 )] = 244.9
                            t = 15.64 s.
                            x =119 m/s (15.64 s) = 1868 m
                            f = tan-1 (1868/1200) = 570 .
Problem 2: A movie stuntman is to run across a rooftop and jump horizontally off it, to land on the roof of the next building. Before doing the stunt, he wisely asks you to determine if it is possible. Can he make the jump if his maximum rooftop speed is 4.5 m/s?
Solution:
                    The fall of 4.8 m will take a time t. Let y – y0 = -4.8 m , negative due to the choice of origin. Since he jumps off HORIZONTALLY, then q0 = 0.
                            y – y0 = 0t – ½ gt2
                            t = [2(y – y0 )/(-9.80 m/s2 )]1/2 = [2(-4.8 m)/(-9.80 m/s2 )]1/2 = 0.99 s
                    We next have to find the x-distance he will cover in 0.99 s:
                            x = 4.5 m/s (cos 0)(0.99 s) = 4.5 m.
                    Since the building is 6.2 m over, your advice should be: DON’T JUMP.
Problem 3: A stone is thrown from the top of a building upward at an angle of 40.00 to the horizontal with an initial speed of 30.0 m/s. The height of the building is 50.0 m.
a) How long does it take for the stone to hit the ground?
Solution:
                    Knowns: yo = 50.0 m, y = 0, vo = 30.0 m/s, and q = 40.0o
                    y = yo +(vosinq)t -(1/2)gt2
                    0 = 50.0  + 19.3 t - 4.9t2
                    We must use the quadratic equation to solve for t.  Using the quadratic equation we obtain 2 answers: t = -1.78 s and t = 5.72 s  and our choice is obvious, the positive value t = 5.72 s.

b) With what velocity does the stone hit the ground?
Solution:
                    The velocity in the x - direction is constant and is equal to:
                            vx = vocosq = (30.0 m/s)(cos40.0o) = 23.0 m/s
                    The velocity in the y - direction changes because there is acceleration in the y -direction:
                            vy = vosinq - gt = (30.0 m/s)(sin40.0o) - (9.8 m/s2)(5.72 s) = -36.8 m/s
                    Therefore:  v = (23.0 m/s)i + (-36.8 m/s)j
                            |v| = [(23.0)2 + (-36.8)2]1/2 = 43.4 m/s
                           q = tan-1(-36.8/23.0) = -58.0o

c) Where does the stone strike the ground (i.e., how far away from the base of the building)?
Solution:

                    x = (vocosq)t = (30.0 m/s)(cos40.0o)(5.72 s) = 131 m
Problem 4: A river has a steady speed of 0.500 m/s. A student swims upstream a distance of 1.00 km and returns to the starting point. If the student can swim at a speed of 1.20 m/s in still water, how long does the trip take? Compare this to the time the trip would take if the water were still.
Solution:
                    The total time in still water t = d/v = 2000/1.2 = 1.67 x 103 s
                    Total time = time upstream + time downstream
                    tup = 1000/(1.20 - 0.500) = 1429 s
                    tdown = 1000/(1.20 + 0.500) = 588s
                    Total time = 1429 s + 588 s  = 2017 s = 2.02 x 103 s