Question

1. A ball is thrown horizontally at
a speed of 40mts at an angle of
60° with the horizontal. Find (a)
(6) the range of
man. height reached and
the reall.
Take a
- 10mg​

Answer 1

Answer:

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

Explanation: