Question

An object is projected at an angle of elevation of 45° with a velocity of 100 m/s.Calculate it's :-
(1) range?
(2) time of flight?

Answer 1

Range =1019xm

T=14.4xs

Explanation:

To get the time of flight T use:

v=u+at

To get the time t to reach the maximum height, take vertical components of motion:

0=vsinθ−gt

∴t=vsinθg=100sin459.81=7.208xm

Since the motion is symmetrical we can say that the total time of flight T will be given by:

T=2t=2×7.208=14.41xs

To get the range use the horizontal component of velocity which is constant.

s=vcosθ×t

∴s=100cos45×14.41=1019xm