Question

A ball is thrown upwards with a velocity 20 ms from the top of tower having a height 50 m. whai is time elasped when it passes again edge of tower​

Answer 1

Answer:

bro my answer is correct

that is 4 seconds

see , velocity is given as 20 m/s

g = -10 m/s2

put these values in v=u+ at

v=0

a=-g=10

0=20-10t

-20=-10t

t=2s

2 seconds will be consumed in going up by the ball and 2 seconds will be consumed by coming down by the ball so the ball will cross the edge after 4 seconds

answer =2+2 = 4 seconds

plz mark as brainliest

Answer 2

Answer:

Heya!!

Step-by-step explanation:

given that,

a ball is thrown upwards with a velocity 20 m/s from the top of tower

here,

initial velocity of the ball = 20 m/s

let the maximum height reached by ball from the initial position be h

so,

at its maximum height velocity will 0

so,

final velocity of the ball = 0 m/s

now we have,

Initial velocity(u) = 20 m/s

final velocity(v) = 0 m/s

gravitational acceleration(g) = -10 m/s²

Now,

by the gravitational equation of motion,

v = u + gt

where,

t is the time taken to reach maximum height

putting the values,

0 = 20 + 2(-10)t

-20t = -20

t = -20/-20

t = 1 s

so,

time taken to reach maximum height

= 1 seconds

now we know that,

Time taken to reach maximum height

= time taken to reach its initial position

so,

time elapsed when it passes again edge of tower = 1 + 1

= 2 seconds

Time taken to pass edge of tower

Time taken to pass edge of tower= 2 seconds