Question

a particle is thrown vertically up with a velocity of 50 what will be it's velocity at highest point of jurney ? how high would the particle rise? what time it takes to rise at the highest point ? (take the value of g = 10 m/s)​

Answer 1

1. Explanation:

at the highest point... velocity becomes 0

because at that time...it starts to turn...

v^2 = 2as

50^2 = 2×10×S

2500÷ 2÷10 = S

s = 125( distance)

v = u + at

0 = 50+ 10T. (gravity is negative)

50÷10= T

t = 5s ( time)

Answer 2

Answer:

its velocity at highest point of journey is 0.

acceleration, a=g=10m/s^2

a=-10

v=0

u=50m/s

$2as = {v}^{2} - {u}^{2} \\ s= {v}^{2} - {u}^{2} \div 2a \\ s = {0}^{2} - {(50)}^{2} \div 2 \times - 10 \\ s = - 2500 \div 2 \times - 10$

s=2500/20

s=125m

the particle goes 125m up

V = U + at

V - U /a=t

0-50/10=t

t=5sec