A ball rolls on a horizontal table and drops. it hits the 0.

[JAZZ541]

Homework Statement

it hits the ground 0.5s after dropping and at a 40cm distance from the table.
what is the table height?
what is the ball initial speed?
in what speed (with direction) did it hit the ground

Homework Equations

The Attempt at a Solution

solved thee first question stuck on the other two
thx for any help!

d = v0t + ½gt2

v0 = 0

t = 0.5s

g = 10m/s2

d = 0 + ½• (10m/s2 • (0.5s)2 = 1.25m

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[BvU]

Hi jazz,

You did well on part 1. What equation describes the horizontal motion during these 0.5 seconds ?

 

[JAZZ541]

hi ty for your reply!
I solved question 2 also here's what I did:

X(t) = X0 + V0xt

X(t) = 40cm = 0.4m

0.4m = 0 + V0x • 0.5s

0.5sV0x = 0.4m

V0 = 0.8m/s

they also ask in what direction, what does that mean?
still struggling on question 3 would appreciate any guidance, thanks again!

 

[jbriggs444]

they also ask in what direction, what does that mean?

The ball will hit the ground at some angle above the horizontal.

 

[BvU]

Good. So the ball has 0.8 m/s horizontal speed. How much vertical speed does it have when it hits the ground ?

 

[JAZZ541]

The ball will hit the ground at some angle above the horizontal.

where?

 

[BvU]

Good thing you made a drawing. If the floor is at the end of the red line, there !

 

[JAZZ541]

Vy(t) = V0y – gt

0 = V0y – 10m/s2 • 0.5s

V0y = 5m/s

is that right?

 

[BvU]

I would say that $\ v_y(0) = 0 \$ as in your first post and therefore $v_y(0.5) = -5$ m/s. The minus sign indicates it is going down.

So you now have horizontal and vertical speeds $\vec v(0.5) = (0.8, -5)$ m/s. Can you calculate the magnitude of the total velocity and the angle with the horizontal ?

 

[JAZZ541]

think I made it:Tanα = 5/0.8 → α = 80.9°

Y = 5sin80.9° = 4.937

X = 0.8cos80.9° = 0.126

4.937 + 0.126 = 5.06m/s

 

[BvU]

I would say $-5/0.8$.

And $v_x = 0.8, \ \ v_y = -5$ which with Pythagoras, gives $v^2 = 25 + 0.64 \ \Rightarrow \ v = 5.06$ m/s

(by the way, 5sin80.9° = 3.28 and 0.8cos80.9° = 0.57 !)

 

[JAZZ541]

haha you make my efforts look so complicated when there's a clean simple answer
and I calculated when calc is on degree, that's how my teacher wants it, I thought I can't use Pythagoras I can't remeber why now, very new to all of this, is what I did legit at all? cause if its is (-) then result would be different

 

[BvU]

What I meant to say is that 5 * sin 80.9° = 4.94 and 0.8 * cos 80.9° = 0.79 ().

And what you meant to say is that 5.063 * sin 80.9° = 5.0 and 5.063 * cos 80.9° = 0.8

That make us even ?

 

[JAZZ541]

haha, too much thinking made me a bit slow I guess, ty for your help! unfortunately I'm sure Ill need it again lol