A projectile is thrown up with the initial speed u, making an angle theta (theta > 45 deg) with the horizontal. What is the time, just after which, it will be moving perpendicular to its initial direction of motion?

a) u/(gsintheta)

b) (usintheta)/g

c) u/(gcostheta)

d) (ucostheta)/g

1 Answer
Dec 19, 2017

THe answer is option (a) that is =u/(gsintheta)

Explanation:

Reminder :

sin^2theta+cos^2theta=1

The initial speed is =u

The angle is =theta,,(theta>45^@)

Initial conditions

The components of the speed are

V_(x_0)=ucostheta

V_(y_0)=usin theta

The components of the speed after time t are

V_(x_t)=ucostheta

V_(y_t)=usintheta- g t, From the equation of motion v=u+at

The acceleration due to gravity is =g

Writing the components of the speed in vector notation

((ucostheta),(usintheta)) and ((ucostheta),(usintheta- g t))

In order for the projectile to move perpendicular to the initial direction, the dot products =0

((ucostheta),(usintheta)).((ucostheta),(usintheta- g t))=0

u^2cos^2theta+u^2sin^2theta-u g tsintheta=0

u^2-ug tsintheta=0

t=u/(gsintheta)