Question

When a body is thrown with a velocity u making an angle θ with the horizontal plane, the maximum distance covered by it in horizontal direction is​

Answer 1

in horizontal direction

velocity is constant as there is no acceleration

v=ucosθ

time of flight=

g

2usinθ

range=maximum distance traveled in horizontal direction=vt=

g

u

2

sin2θ

Hope it helps.

Answer 2

Answer:

Explanation:

u=  ucosθ  i  +usinθ  j

ux    = constant

ux    (t)=ucosθ

u  y  (t)=usinθ−gt

So, u(t)=ucosθ  i  +(usinθ−gt)  j

Now, for u(t) to be perpendicular to u, dot product should be  0.

u(t) ⋅u  =0

(ucosθ  i +(usinθ−gt )j )⋅(ucosθ  i  +usinθ  j  )=0

u ^2 cos  ^2 θ+u  ^2 .sin ^2 θ−usinθgt=0

t=  gsinθ /u

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