Question

1. A ball is thrown up with a speed of 0.5 m/s . (i) How high will it go before it begins to fall ? (ii) How long will it take to reach that height ? pls make it fast

Answer 1

Explanation:

A given :-

u = 0.5 m/s

g = 9.8 m/s^2 ( it is taken to be negative since body is going against the gravity .

v = 0

s = ?

v^2 - u^2 = 2gs

(0)^2 - (0.5)^2 = 2(-9.8) S

- (0.5 * 0.5 ) = -2 * 9.8 * S

S = 0.5 *0.5 / 2 * 9.8 = 0.012 m is answer

B V=u+gt

V=0m/a

U=0.5m/s

g=-10

0=0.5-(10.t)

-0.5=-10t

t=0.05

Answer 2

Explanation:

given

v=0

u=.5

A= -10M/s²

1 we have to find distance or max hight or s

by 3rd equation of motion

v²= u²-2at

there is negative singn coz obejct in goinng up wards against acceleration

by putting value

0²=.5²- 2(-10)s

0= (2.5)+20s

0= 2.5 + 20s

2.5= 20s

s=2.5/2x10

s = .125m

2 time

so speed = distance / time

then time = distance/speed

so time = .125/.5

time = .25 second