Question

A particle is projected upwards with a velocity of 100 m//sec at an angle of 60^(@) with the vertical. Find the time when the particle will move perpendicular to its initial direction, taking g = 10m//sec^(2).

Answer 1

Answer:

Your answer is 20 seconds. It can be found out by-

1- resolving the velocity vector into horizontal (50√3 m/sec) and vertical components (50 m/sec)

2-finding the time required to reach the maximum height(5 sec)

3-representing the vector which is perpendicular to the initial direction as a ray which is inclined at an angle of 60° below the horizontal (which is represented by the horizontal component of the initial velocity vector-50√3 m/sec)

4-finding the corresponding vertical component(150 m/sec)

5-finding the time taken for the velocity to reach 150 m/sec from the maximum height(15 sec)

6-adding the time found in step 2 with the time found in step 5