Answer:
(a) 2.23 m/s
(b) 0.9 m
Explanation:
h = 1 m, t = 0.1 second
horizontal component of initial velocity, ux = 2 m/s
vertical component of initial velocity, uy = 0
(a) Let v be the velocity after 0.1 seconds. Its vertical component is vy and horizontal component is vx.
The horizontal component of velocity remains constant as in this direction, acceleration is zero.
vx = ux = 2 m/s
Use first equation of motion in Y axis direction.
vy = uy + g t
vy = 0 + 9.8 x 0.1 = 0.98 m/s
Resultant velocity after 0.1 second
v^2 = vx^2 + vy^2
v^2 = 2^2 + 0.98^2
v = 2.23 m/s
(b) Let it takes time t to land.
Use second equation of motion along Y axis
h = uy t + 1/2 g t^2
1 = 0 + 1/2 x 9.8 x t^2
t = 0.45 second
Let it lan at a distance x.
so, x = ux x t
x = 2 x 0.45 = 0.9 m
