Question

G A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 15.2 m/sm/s .

A) What is its speed after falling for 2.30s (in m/s)?

B) How far does it fall in 2.30s (in meters)?

C) What is the magnitude of its velocity after falling 11.0m (in m/s)?

Answer 1

Step-by-step explanation:

Using the equations of motion,

i. v = u + gt

ii. S = ut + (gt²)/2

iii. v² = u² + 2gS

Where,

g = 9.81 m/s²

Initial velocity, u = 15.2 m/s,

t =2.3 z

A.

Using i.,

v = 15.2 + (9.81 ×2.3)

= 37.8 m/s

B.

Using ii.,

S = (15.2×2.3) + 9.81 * (2.3²)/2

y = 60.9 m

C.

Using iii.,

v² = 15.2² + (2×9.8×11)

v² = 446.86

= 21.1 m/s

Answer 2

a) Speed after falling for 2.30s = 37.74 m/s

b) The water ballon falls 60.88m in 2.3 s

c) Velocity after falling 11.0m = 21.13 m/s

Step-by-step explanation:

Using the equations of motion,

g = 9.8 m/s²

Initial velocity, u = 15.2 m/s,

velocity at t=2.3s, v = ?

a) v = u + gt

v = 15.2 + 9.8×2.3

v = 37.74 m/s

b) y = ut + gt²/2

y = (15.2×2.3) + 9.8(2.3²)/2

y = 60.88 m

c) v² = u² + 2gy

v² = 15.2² + (2×9.8×11)

v = 21.13 m/s