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?

Question

a) #u/(gsintheta)#

b) #(usintheta)/g#

c) #u/(gcostheta)#

d) #(ucostheta)/g#

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Answer 1 · 2017-12-19T12:53:17

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

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)#