H.W: An object is thrown horizontally at 20m/s from a cliff top 45m above sea level. Find (a)the time to land; (b)the angle at which it hits the water; (c)the speed at which it hits the water. Ans (al303 54 73 (e)3S 79m/s
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- 10. A projectile is launched at t = 0.0 s from the edge of a cliff at a height (h) of 100 m above the water, with a velocity vo-100 m/s, O=60.0° above the horizontal. The projectile rises and falls into the sea. Determine the time it takes the projectile to reach the highest point in its trajectory, and the speed of the projectile at this point:e Time (s) Speed (m/s)< A) 5.10 100 В) 8.84 h C) 8.84 50.0 D) 17.7 86.6 E) 17.7A group of hikers sets out going 10.0km [N], then 15.0km [E], then 4.0km [S]. The entire trip takes 4.00h. Determine the magnitude hikers' velocity for the trip. 4.04 km/h [N 21.8° E] 5.25 km/h [N 21.8° E] 7.25 km/h [N 21.8° E] 21.0 km/h [N 21.8° E] 65.25 km/h [N 21.8° E]1.A conductor in a train travelling at 4.0 m/s (N) walks across the train at 1.2 m/s (E) to validate a ticket. If the trian car is 4.0m wide, how long does it take the conductor to reach the other side? vog=? vom 1.2 m/s(E) (1.2/4.2) vmg= 4.0 m/s(N) vog 4.2m/s(N17E) Ad At = Δυ At = 3.3s vom=5m/s(W) vmg=11m/s(S = 12m/s = vog vom+vmg -5m/s (W)+ 11m/s(S) = sqrt (5m/s)^2+ (1m/s)^2 sin 90 12m/s 4.0m/s 1.2m/s 2. A jetski speeds across a river at 11 m/s relative to the water. The jet ski's heading is South. The river is flowing West at 5m/s. Determine the jet ski's velocity relative to the shore. vog=? = vog vom+vmg sin 0 5m/s = 1.2m/s(E) + 4.0m/s(N) =sqrt (1.2m/s)^2 + (4.0m/s)^2 =4.2 m/s = 24° Therefore the answer is 12m/s (S24W) Tan-1= = 17
- The minimum speed he'd need to run off the edge of the cliff to make it safely to the far side of the river?Starting from the relative position shown, aircraft B is to rendezvous with the refueling tanker A. If B is to arrive in close proximity to A in a 2.3-minute time interval, what absolute velocity vector should B acquire and maintain? The velocity of the tanker A is 306 mi/hr along the constant-altitude path shown. 8960 A 2710' В Answer: VB = ( i i+ i j) mi/hrJse the worked example above to help you solve this problem. An Alaskan rescue plane drops a package of emergency rations to stranded hikers, as shown in the figure. The plane is traveling horizontally at 45.0 m/s at a height of 1.40 x 104 m above the ground. (a) Where does the package strike the ground relative to the point at which it was released? 241 (b) What are the horizontal and vertical components of the velocity of the package just before it hits the ground? horizontal 45 m/s 52.3832 vertical The response you submitted has the wrong sign. m/s (c) Find the angle of the impact. 49 The response you submitted has the wrong sign.°
- 3. The graph below shows the displacement vs time of a student leaving the NSCI 110 after physics class walking towards the cafeteria for lunch (assume a 1-D motion) At some point between NSCI 110 and the cafeteria, the student realizes they have left their cell phone in the classroom, so they turn back to get it. Displacement vs Time 120 100 nol Velocity vs Time 80 1.5 60 40 30 60 90 120 150 180 210 20 -1.5 30 60 90 120 150 180 210 Time (s) a. How far did the student walk before realizing that they left their cell phone in the classroom? How long did this take? (2 marks) b. With what velocity did the student walk back to the classroom? (2 marks) c. Fill out the velocity vs time graph above. (2 marks) Distance (m)Bob has just finished climbing a sheer cliff above a level beach and wants to figure out how high he climbed. All he has to use is a baseball, a stopwatch, and a friend on the ground below with a long measuring tape. Bob is a pitcher and knows that the fastest he can throw the ball is about vo = 33.7 m/s. Bob starts the stopwatch as he throws the ball, with no way to measure the ball's initial trajectory, and watches carefully. The ball rises and then falls, and after t₁ = 0.710 s the ball is once again level with Bob. Bob cannot see well enough to time when the ball hits the ground. Bob's friend then measures that the ball hit the ground x = 123 m from the base of the cliff. How high above the beach was the ball when it was thrown? initial height: X 1₁ mNASA launches a rocket at t = is given by h(t) = - 4.912 + 349t + 318. O seconds. Its height, in meters above sea-level, as a function of time Assuming that the rocket will splash down into the ocean, at what time does splashdown occur? The rocket splashes down after seconds. How high above sea-level does the rocket get at its peak? The rocket peaks at meters above sea-level.