Solution :
initial velocity=u=40 m/s
a=g=9.8m/s2
angle of projection=θ=60°
A) maximum Height h= u²sin²θ/2g
h= 40x40 sin60²/2x10
h= 40x40 x3/4x2 x 10
h=60m
B) Horizontal Range = X=u²sin2θ/g
=X= 40x40 sin2x60/10
=40x40 x√3/2x10
=80√3 m
Answer:
c) x = 200 m
Explanation:
Ball is projected at speed of 40 m/s at an angle of 60 degree
so here we have two components of velocity given as
[tex]v_x = 40 cos60[/tex]
[tex]v_x = 20 m/s[/tex]
[tex]v_y = 40 sin60[/tex]
[tex]v_y = 34.64 m/s[/tex]
now the horizontal distance moved by the ball in next 10 s is given as
[tex]x = v_x \times t[/tex]
so we have
[tex]x = 20 (10)[/tex]
[tex]x = 200 m[/tex]
