Question

2. 1-D Kinematics, Constant Acceleration After falling a distance of 45.0 m from the top of a building, a box is landing on the top of a car, which is crushed to a depth of 0.50 m due to the impact. g = 9.8m/s2 Calculate: a) the speed of the box just before it touches the top of the car. b) The average acceleration of the box during the impact. c) The time it took to crush the roof.

3. 2-D Kinematics An octopus swimming in a horizontal plane has the velocity ~v = (4.0ˆx + 1.00ˆy)m/s at the time t = 0. After the octopus swims for 10.0 s with a constant acceleration, its velocity is ~v = (20.0ˆx − 5.00ˆy)m/s. a) What are the components of the acceleration? b) What is the direction of the acceleration relative to the unit vector ˆx?

4. 2-D Kinematics An archer shoots an arrow with a velocity of 50.0m/s at an angle of 45 deg with the horizon. An assistant standing on the level ground 150 m downrange from the launch point throws an apple straight up. g = 9.8m/s2 a) What is the height of the arrow after it has traveled a horizontal distance of 150 m? b) What is the minimal launch speed of the apple to meet the path of the apple? c) At what time after the arrow launch should the apple be thrown so that the arrow hits the apple?

5. 2-D Kinematics For a projectile that lands at the same height as it is released, show that the maximum horizontal distance is reached if the projectile is released at an angle of 45 deg with respect to the horizontal.

Answer 1

2.

a)

consider the motion from the top of building to the point just above the top of car

v_o = initial velocity at the top of building = 0 m/s

v_f= final velocity just before hitting the top of car = ?

a = acceleration = acceleration due to gravity = 9.8 m/s²

Y = displacement = 45 m

using the kinematics equation

v_f² = v_o² + 2 a Y

v_f² = 0² + 2 (9.8) (45)

v_f = 29.7 m/s

b)

Consider the motion during the impact :

v_o = initial velocity just before hitting the top of car = 29.7 m/s

v_f= final velocity = 0 m/s

d = stopping distance = 0.50 m

a = acceleration during impact

using the kinematics equation

v_f² = v_o² + 2 a d

0² = 29.7² + 2 a (0.50)

a = - 882.1 m/s²

c)

t = time taken to crush the roof

Using the equation

v_f = v_o + a t

0 = 29.7 + (- 882.1) t

t = 0.034 sec