Projectile Motion Homework: Finding Height and Maximum Height - Physics

[rissie]

Homework Statement

A ball is thrown with a velocity of 20.0 m/s [30 degrees] and travels for 3.0 s before it strikes the ground. Find the
a) height from which it was thrown.
b) the maximum height of the ball.

Homework Equations

a) d_x = v_x • t ... I don't even know what to do for part A.
b) v²_f = v²_i + 2ad

The Attempt at a Solution

a) I absolutely have no clue what to do for part A.

I first thought to use triangles and tried to find the distance the ball traveled using the velocity I found, but it doesn't even work.

b) I realized that at the maximum height, the velocity of the object will me 0 m/s, so I tried to find the distance from that point to the ground to find the maximum height.

I found the initial velocity to be 10 m/s using SOH and i used the above formula.
v_iy: 10 m/s
v_fy: 0 m/s
g(a): -9.81 m/s²

0² m/s= 10² m/s + 2 • -9.81 m/s² • d
0 m/s = 100 m/s + -19.62 m/s² • d
-100 m/s = -19.62 m/s² • d
d = 5.1 m

It's not the right answer, so I don't know where I went wrong.

I don't know if any of what I did makes sense, I'm so lost in projectile motion.

 

[rl.bhat]

Hi rissie, welcome to PF.

When the ball reaches the ground, the displacement of the ball is d, the height from which the ball is thrown. Initial velocity is 10 m/s. The total time is 3 s.
Use the formula
-d = V(iy)*t - 1/2*g*t^2.

Substitute the values and find d.

 

[rissie]

Ohhh, okay. That makes so much more sense now.

That's just for the height it was launched from, right?

 

[rl.bhat]

Yes. For maximum height add 5.1 m to d.