Question

plz solve this question its urgent and dont spam it​

Answer 1

time taken is 5sec, velocity of object is 5m/s

avarage speed=total distance cover/total time taken.....................(a)

here distance cover along horizontal and vertical is,

x=ut+1/2at²............ (along horizontal)

here a is 0 because friction due to air along horizontal seem to zero.so there is no change in acceleration it remains constant or zero.

x=ut. (or) t=x/u..............(1)

x=5*5=25m

x=ut+1/2at²............(along vertical)

here friction of air apply so there is some change in accelration which is equal to accelration due to gravity(g), but here it's vertical velocity is zero.

y=0t+1/2gt²

y=1/2gt²...............(2)

from equation (1)

y=1/2g(x/u)²

$y = ( \frac{g}{2 {u}^{2} } ) {x}^{2}$

y=(10/2*5²)25²

y=(10/50)625

y=0.2*625

y=125

time taken is, t=y/u=125/5=25 sec

from equation (a)

avarage speed=

(2)25+125/5+25=

2*6.67m/s

13.4m/s