Question

A ball is thrown at a speed of 40 m/s at an angle of 60° with the horizontal. Find (a) the maximum height reached and(b) the range of the ball. Take g=10 m/s².
Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

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 2

HEY!!

_____________________________

✔Initial speed (u) = 40 m/s

✔Angle of projection of the ball with the horizontal(α) = 60°

✔a = g = 10 m/s^2

a) Height reached by the ball=

H=u^2 sin^2 α/2g
 
▶▶H=40^2  (sin 60°)^2/ 2×10=60 m

b) Horizontal range of the ball=

R=u^2sin2α/g  

40^2 sin(2×60°)/10= 80 root 3 bar m.