Question

A B C D 14. A steel ball is dropped from a diving platform (with an initial velocity of zero). Using the approximate value of g = 10 m/s^2, what is its velocity 2.4 seconds after its release? a) 10, b) 2.4, c) 24, d) 4.2

A B C D 15. A large rock is dropped from the top of a high cliff. Assuming that air resistance can be ignored and that the acceleration has the constant value of 10 m/s^2. how fast would the rock be traveling 8 seconds after it is dropped in m/s? a) 40, b) 1.25, c) 18, d) 2.

A B C D 16. A ball is dropped from a high building. Using the approximate value of g = 10 m/s^2. find the change in velocity between the first and

fourth second of its flight in m/s. a) 30, b) 3.33, c) 13, d) 40

A B C D 17. A ball is thrown upward with an initial velocity of 12 m/s. Using the approximate value of g = 10 m/s^2, how high above the ground is the ball 2 seconds after it is thrown in meters? a) 12, b) 24, c) 6, d) 4.

A B C D 18. Suppose that the gravitational acceleration on a certain planet is only 3.0 m/s^2. A space explorer standing on this planet throws a ball straight upward with an initial velocity of 18 m/s. How much time in seconds elapses before the ball reaches the high point in its flight? a) 3, b) 6, c) 18, d) 54.

A B C D 19. A ball rolls off a shelf with a horizontal velocity of 5 m/s. At what horizontal distance in meters from the shelf does the ball land if it takes 0.4 s to reach the floor? a) 5.4, b) 2, c) 4.6, d) 1.25.

Answer 1

14. option c) 24 is correct here

here we have given that

g = 10 m/s^2,

so that velocity 2.4 seconds after its release

v = u + gt = 0+ 10*2.4 =24 m/s

15.

here we have given that

acceleration has a constant value of 10 m/s^2.

for how fast would the rock be traveling 8 seconds after it is dropped in m/s

so that

v = u + gt = 0 + 10*8 = 80 m/s

so none of the given option is correct here

16

here we have given that

the approximate value of g = 10 m/s^2.

the change in velocity between the first and fourth second of its flight in m/s. will be given as

delta v = delta u + g * delta T = 0 + 10 * 3 = 30 m/s

hence option  a) 30, is correct here

17.

here we have given that

initial velocity of 12 m/s

for how high above the ground is the ball 2 seconds after it is thrown in meters

we have

S = ut - 0.5 gt^2 = 12*2- 0.5*10*4= 4 m

hence option d) 4m is correct here

18.

acceleration on a certain planet is only 3.0 m/s^2

initial velocity of 18 m/s.

For how much time in seconds elapses before the ball reaches the high point in its flight we have

here the final velocity becomes zero at the highest point so that

v = u -gt

u/g = t

hence t = 18/3=6

hence option b) 6 is correct here

19.

here we have given that,

horizontal velocity of 5 m/s

it takes 0.4 s to reach the floor
so that here

here we have since in  the horizontal direction, there is no acceleration so we can simply use

velocity=S/t
Hence distance along the floor from the table is 2×0.4=2m

hence here option b)2 is correct .