Assignment Instructions Complete the following questions fro

Assignment Instructions: Complete the following questions from the textbook, showing all work clearly, including assumptions.

1. A person walks 20.0 degrees south of east for 12 km. How for would the need to walk due south and due east to arrive at the same location?

2. A boat crosses a .26 km wide river flows with a constant speed of 8.0 m/s to the south. It takes a boat 53 seconds to cross the river to a point directly west of the departure point of the boat. What is the speed of the boat relative to the water?

3. A ball is thrown straight upwards and is caught by a person on a a balcony when the ball is at its highest point, which was 19.0 m above where the ball was released. What was the initial velocity of the ball?

4. An arrow is aimed horizontally at the center of a target which is 30.0 m away. The initial speed of the arrow is 78.0 m/s. What is the distance from the center of the target to the point where the arrow strikes the target?

5. A ball with a mass of 5.0 kg is dropped from the top of a building that is 53 m tall. Derive the expression that describes the height of the building as a function of time after the ball is released. Determine how long it takes for the ball to strike the ground.

Solution

1. Due South would be 12 * sin 20 = 12*.342 = 4.10 Km

   Due East would be 12 * cos 20 = 11.268 KM.

2. vertical velocity of boat relative to boat =0.

horizontal velocity of boat = .26*1000/53 =4.91 m/s

Relative Velocity = (4.91*4.91 + 8*8)^1/2 = 9.386 m/s

Angle = tan^-1 4.91/8 = 31.5 degree approx north of west.

3. velocity = (2gh)^1/2.

So initial velocity = (2*9.8*19)^1/2 = 18.49 m/s .

4. Time taken to reach target ,t₁= 30/78 s.

Distance travelled downwards due to gravity = g/t₁ = 9.81/(30/78) = 25.5 m.

5. Initial velocity of Ball = 0. Accelaration =gravity = 9.81 m/s

Let time taken =t.

height of building= ut + 1/2 at² = 1/2 *9.81 *t²

height of building =4.95t^{2 .}

time taken to reach ground = (53/4.95)^1/2 = 3.27 sec.