Question

A body thrown vertically upward from the ground with a velocity v reaches a maximum height h. If it is thrown at an angle of 45o, with what velocity should it be thrown to reach the same maximum height? In this case at what horizontal distance would it fall to ground. (Hint: to reach same vertical distance, the vertical component of velocity should be same as v. Also, obtain the range in terms of h)

Answer 1

Answer:

the velocity should be

$\frac{v}{ \sqrt{2} }$

the horizontal distance should be

$\frac{ {v}^{2} }{2g}$

Explanation:

Here,

the volocity must be ....

$v \sin(45)$

and the horizontal distance....

$x = \frac{ {v}^{2} \times ( { \sin(45)) }^{2} \times \sin(90) }{g}$

which gets to....

$x = \frac{ {v}^{2} }{2g}$

I am done.