A ball is thrown directly downward with an initial speed of 8.00 m/s, from a hei

Question

A ball is thrown directly downward with an initial speed of 8.00 m/s, from a height of 30.0 m. After what time interval does it strike the ground? A certain freely falling object, released from rest, requires 1.50 s to travel the last 30.0 m before it hits the ground. (a) Find the velocity of the object when it is 30.0 m above the ground. (a) Find the total distance the object travels during the fall. An attacker at the base of a castle wall 3.65 m high throws a rock straight up with speed 7.40 m/s at a height of 1.55 m above the ground. (a) Will the rock reach the top of the wall? (b) If so, what is the rock's speed at the top? If not, what initial speed must the rock have to reach the top? (c) Find the change in the speed of a rock thrown straight down from the top of the wall at an initial speed of 7.40 m/s and moving between the same two points. (d) Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations? Explain physically why or why not.

Explanation / Answer

46. Applying vf^2 - vi^2 = 2 a d

v^2 - 8^2 = 2 x 9.8 x 30

v = 25.5 m/s

47. Applying d = v0 t + a t^2 /2

30 = (1.50 v) + (9.8 x 1.50^2 / 2 )

v = 27.35 m/s ....this is the speed at the starting of last 1.50 sec.

(a) v = 27.35 m/s

(b) Aplying vf^2 - vi^2 = 2 a d

27.35^2 - 0^2 = 2 x 9.8 x d

d = 38.2 m

total distance travelled =38.2 + 30 = 68.2 m

48. (A) for maximum height.

0^2 - 7.40^2 = 2 x -9.8 x h

h = 2.79 m

maximum height from ground = 2.79 + 1.55 = 4.34 m

(B) v^2 - 7.40^2 = 2(-9.8)(3.65 - 1.55)

v = 3.69 m/s

(c) v^2 - 7.40^2 = 2(-9.8)(-2.79)

v = 10.5 m/s

change in speed = 10.5 - 7.40 =3.1 m/s

(D) change will not be same because acceleration is same in both cases but time taken will be different on both cases.

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