Question

A body of mass m is thrown upwards at an angle  with the horizontal with velocity v.

While rising up the velocity of the mass after tseconds will be​

Answer 1

It will become zero at the peak and then decreases

Answer 2

Given :-

A body of mass m is thrown at angle.

Let the angle = Ø.

The horizontal component of Velocity = Vx

The vertical component of Velocity = Vy

|V(t)| = √(Vx)² + (Vy)²

Vx = V CosØ

Vx = V CosØVy = V SinØ - gt

|V(t)| = √(V CosØ)² + (V SinØ - gt)²

|V(t)| = √V² Cos²Ø + V² Sin²Ø + g²t² - 2V SinØ gt

|V(t)| = √V²(Cos²Ø + Sin²Ø) + g²t² - 2V SinØ gt

|V(t)| = √V² + g²t² - 2V SinØ gt.

Answer......