Problem with horizontal projectile motion questions

Question

I don't really understand horizontal projectile motion. Suppose I throw a ball horizontally. Given $\theta$ = 0 degrees, time = 0.5 seconds, horizontal range = 15.5 metres,

the "initial horizontal velocity of the ball" would just be the average velocity which is displacement divided by time? Or is it $V\sin\theta$?

Also, what would the "final velocity" be? It would be $\sqrt{V_x^2 + V_y^2}$ using the final values since $\theta$ changes?

And would I say that the maximum height reached by the ball is 0 metres or -1.2 metres?

Finally, in order for me to calculate Kinetic Energy, Work, Force, would I need to measure the mass of the ball or can I get the mass using the information I already know such as velocity and displacement and time?

Answer 1

At time 0 when you throw the ball, there only exists the horizontal velocity you gave it, as the acceleration due to gravity hasn't created a vertical velocity yet. This horizontal velocity will remain constant until the ball hits the ground and eventually stops. Gravity is adding a perpendicular component that will affect the resultant velocity, but not the horizontal component you gave it.

Knowing the time it takes for the ball to hit the ground and the distance travelled it is possible to obtain the initial horizontal velocity, simply by calculating it as the horizontal distance travelled divided by the time it was airborne, as this component of the velocity is the only one driving the ball forward: vh * 0.5s = 15.5m -> vh = 31m/s

The final velocity would be calculated as you propose, vh would be 31m/s and vy can be found knowing the acceleration due to gravity and the time elapsed.

How you define your maximum height is a matter of notation.

You would need the mass of the object in order to calculate kinetic energy, forces and work.

Answer 2

Assuming no air resistance the horizontal velocity will not change. You can calculate the horizontal velocity from the given time and distance.

Vertically is an accelerated motion with constant acceleration. You can calculate the initial height from there.

You are right about the final velocity. You just need to calculate $v_y$.

You need the mass for kinetic energy. You can not find the mass from the given information as it applies to all objects when there is no air resistance.

Answer 3

It actually seems confusing. But it all depends on the same factors. The initial height, velocity & g. Let height = h. t = √(2h/g) And distance travelled is V×t as theta is 0.

All velocity you have given is horizontal. No vertical component initially exists.