2. (a) With what speed must a ball be thrown vertically from the ground level to
ID: 1730451 • Letter: 2
Question
2. (a) With what speed must a ball be thrown vertically from the ground level to rise to a maximum height of 50m? (b) How long is the total time of flight? 3. You throw a ball with a speed of 25 m/s at an angle of 42.0° above the horizontal directly toward a wall. The wall is 22 m from the release point o above the release point does the ball hit the wall? (c) What are the When it hits, has it passed the highest point of its trajectory? the ball. (a) How long does the ball take to reach the wall? (b) How far horizontal and vertical components of its velocity as it hits the wall? (d) 4. Write the relationship between the force on a particle and the change potential energy as the particle moves between two points 5 A crate of mass 10kg is pulled up a rough incline with an initial speed of m/s. The pulling force is 100N parallel to the incline, which makes an angle of 20° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5m. (a) How much work is done by gravity? (b) How much energy is lost due to friction? (c) How much work is done by the s 100N force? (d) What is the change in the kinetic energy of the crate? (e) What is the speed of the crate after being pulled 5m? 6. In an arcade game, a spot is programmed to move across the screen according to x-9.00t -0.75t3, where x is distance in centimeters measured from the left edge of the screen and t is time in seconds. When the spot reaches a screen edge, at either x-0 or x-15 cm, t is reset to 0 and the spot starts moving again according to x(t). (a) At what time after starting is the spot instantaneously at rest? (b) Where does this occur? (c) What is the acceleration when this occurs? (d) In what direction is it moving just prior to coming to rest? (e) Just after? () When does it first reach an edge of the screen after t-0?Explanation / Answer
2a) Using third equation of motion,
u^2 = 2gh
u = sqrt(2gh)
= sqrt(2*9.8*50)
= 31.3 m/s answer
b] T = 2 u /g
= 2*31.3/9.8
= 6.39 s
3a] t = d/v cos theta = 22/[25 cos 42 degree] = 1.184 s
b] h = v sin theta t - 0.5 gt^2 = [25 sin 42 degree]*1.184 - 4.9*1.184^2
= 12.94 m
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