Question

answrr fast
14,15 with explaination

Answer 1

Heya user !!

Here's the answer you are looking for

Please see the attachment for question no. 14

Answer of 15th question is here ⬇️

☛ For the downward journey,

Initial velocity ( u' ) = 0 (as it is dropped)

Height (s) = 20m

Acceleration = g = 10m/s

So, let final velocity = u

From 3rd equation of motion,
☞ v² - u² = 2as

Here, u² - (u')² = 2gs

u² - 0 = 2(10)(20)

u = √400 = 20m/s ▶️▶️▶️1️⃣

☛For upward journey,

final velocity ( v' ) = 0

height ( h) = 5m

acceleration = -g = -10m/s²

So, let initial velocity = v

Similarly using 3rd equation of motion,

(v')² - v² = 2(-g)(h)

0 - v² = 2(-10)(5)

v² = 100

v = √100 = 10m/s ▶️▶️▶️2️⃣


⏭️ From 1️⃣ and 2️⃣,

magnitude of change of momentum

= mv - mu

= m ( v - u )

= 1 {10 - (-20)} (-ve sign as direction is different)

= 30kgm/s (answer)



★★ HOPE THAT HELPS ☺️ ★★