Question

2.A body is thrown with velocity of 40 m/s in a direction making an angle of 30° with the
horizontal. Calculate 1) Horizontal range, 2) Maximum height and 3) Time taken to reach the
maximum height.
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Answer 1

Answer:

ok you should have to know is that the time taken and the height only calculated by using the vertical velocity.but the range is calculated by the horizontal component of the velocity.

Explanation:

velocity in component form(34.64,20)

1. max height=

v^2-u^2=2as but at max height the velocity is zero so

v^2=2as

s=v^2/2a a is acceleration due to gravity

s=20^2/2*9.81

s=20.387m

the time it take for half the path is

v=u+at but u=0

v=at so t=v/a

=20/9.81

=2.0387s

so for the whole path we multiply it by 2

2.0387s*2=4.077s

• the range is

v=s/t

s=vt

=34.64*4.077

=141.23m