Question

A small dense ball with mass 1.5 kg is thrown with initial velocity <3, 6, 0> m/s at time t - 0 from a location we choose to call the origin (< 0, 0, 0 > m). Air resistance is negligible. When the ball reaches its maximum height, what is its velocity?

Answer 1

Answer:

y = 1.84 m ,   v= ( 3, 0 ,0) m/s

Explanation:

This is a projectile launching exercise, at the point of maximum height the vertical speed is zero

                $v_{y}$² =  $v_{oy}$² - 2 g y

                 $v_{y}$ = 0

                 y =  $v_{oy}$² / 2g

The components of the initial velocity are

                  v₀ₓ = 3 m / s

                   $v_{oy}$ = 6 m / s

Let's calculate

               y = 6 2/2 9.8

               y = 1.84 m

At this point the speed is only horizontal

               v = vox = 3 m / s

               v = 3i  m/s

               v= ( 3, 0 ,0) m/s

Answer 2

The velocity Vector when the ball reaches its maximum height is;

v = 3i^

Velocity Vector

We are given;

• Mass of ball; m = 1.5 kg

• Initial velocity; u = (3i^ + 6j^) m/s

• Initial position of the ball; x_i = (0, 0, 0)

Now, when the ball reaches the maximum height, the vertical component of the velocity will be zero and thus will have only the horizontal component of velocity.

Therefore in vector form, the velocity vector can be expressed as;

v = 3i^

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