Respuesta :
Answer:
As the projectile starts it’s motion from a vertical height at y=10m, it’s vertical motion may be described as under-
y= 10+ut-0.5gt^2
Where “u” is initial vertical velocity
= 45 sin60,
“g” is acceleration due to the gravity (10m/sec^2)
and “t” is time.
When the projectile touches the ground the vertical height “y”=0
Therefore the time “t” taken for the flight of the projectile till it touches the ground may be calculated from the above equation when y=0 as under -
0= 10+45sin60 x t- 0.5x10 t^2
0= 10+45x (√3/2)x t-5 t^2
t^2-9x (√3/2)x t-2 =0
t= [4.5√3 + √(243/4+4)]/2
t=[7.794+ 8.047]/2
t= 7.92 sec.
The horizontal distance travelled during this time=
vt
Where “v” is horizontal velocity = 45 cos60.
Therefore,
vt =45cos60 x 7.92
= 45x0.5x7.92
= 178.2 m
