Projectile motion above ground given only angle and height

[raptor13]

Homework Statement

I'm having trouble with part A of this question

1. A ball is fired out at an angle of 30° from the platform of height h = 0.5 m as shown in Figure 1. It is subjected to the gravitational force only thus follows a projectile motion. It falls onto the ground after 0.6 seconds. Determine:
a). the ball's initial velocity V₀ when it is fired out.
b). the distance the ball travels in the horizontal direction when it falls on the ground.
c). the magnitude and direction of the ball's velocity at t = 0.3 seconds after being fired out.

Homework Equations

s=((u+v)/2)t
a=(v-u)/t
v=u+at
s=ut+0.5(at ² )
v ² = u² + 2as

where s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time

The Attempt at a Solution

i my first thought was at half the time the final vertical velocity would be zero but then i thought that would only be the case for projectiles that land at the same height as they were fired

 

[haruspex]

but then i thought that would only be the case for projectiles that land at the same height as they were fired

Right. So what is your next attempt?

 

[mfb]

A final vertical velocity of zero would mean the ball travels horizontally. It doesn't do that at the height where it was fired, and it doesn't do that below that height either.
How long does the ball fly upwards? How long does the ball fly downwards?

 

[berkeman]

Homework Statement

I'm having trouble with part A of this question

1. A ball is fired out at an angle of 30° from the platform of height h = 0.5 m as shown in Figure 1. It is subjected to the gravitational force only thus follows a projectile motion. It falls onto the ground after 0.6 seconds. Determine:
a). the ball's initial velocity V₀ when it is fired out.
b). the distance the ball travels in the horizontal direction when it falls on the ground.
c). the magnitude and direction of the ball's velocity at t = 0.3 seconds after being fired out.

Homework Equations

s=((u+v)/2)t
a=(v-u)/t
v=u+at
s=ut+0.5(at ² )
v ² = u² + 2as

where s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time

The Attempt at a Solution

i my first thought was at half the time the final vertical velocity would be zero but then i thought that would only be the case for projectiles that land at the same height as they were fired

Welcome to the PF.

You've listed some of the Relevant Equations, but I think you are missing a couple that will help.

List the equations for the position as a function of time in the vertical and horizontal directions, based on the initial position (x,y) and initial velocities, etc. Then solve the equations.

The horizontal motion will involve the horizontal velocity, which is not changing when there is no air resistance. The vertical motion is affected by the acceleration of gravity.

Can you list the equations for the vertical and horizontal positions as a function of time?EDIT --- Dang, I'm too slow!