Question

a ball is kicked into the
air with an initial vertical speed of 4.9 m/s and a horizontal speed of 6.25 m/s. How long will it remain in the air? What will be its range?

Answer 1

it will remain in air for 1s
and the range will equal to 6.25m

Answer 2

tony115 can you show me the solution?

Answer 3

using the eqn \[v=u+at\] the time required for reaching the maximum height is .5 s same time interval will be taken for reaching back the ground so totally 1s
during this time the ball is also moving along horizontal direction then range
\[r=ut\]
ie 6.5*1=6.5m

Answer 4

what does the "u" stand for?

Answer 5

the initial velociy along vertical direction @jaojao

Answer 6

Can you solve it for me,cause I really can't understand.

Answer 7

Now we have initial speed of 4.9 m/s in the vertical direction @ max. height Velocity will reach zero
so
V= Vo + g * t
g= 9.8 m/s/s
V=0
Vo = 4.9 m/s
Vo = g * t
t= Vo/g = 0.5 Seconds
So time till height is 0.5 s and time till falling again is 0.5 S so total time in air is 1 s
Range = V Horizontal * time = 6.25 m/s * 1 s = 6.25 m