Verifying some confusions :)
Billy-Joe stands on the Talahatchee Bridge kicking stones into the water below.
a) If Billy-Joe kicks a stone with a horizontal velocity of 3.50 m/s, and it lands in the water a horizontal distance of 5.40 m from where Billy-Joe is standing, what is the height of the bridge? (Assume the stone does not bounce on the water.)
b) If the stone had been kicked harder, how would this affect the time it would take to fall?

Question

Verifying some confusions :)
Billy-Joe stands on the Talahatchee Bridge kicking stones into the water below.
a) If Billy-Joe kicks a stone with a horizontal velocity of 3.50 m/s, and it lands in the water a horizontal distance of 5.40 m from where Billy-Joe is standing, what is the height of the bridge? (Assume the stone does not bounce on the water.)
b) If the stone had been kicked harder, how would this affect the time it would take to fall?

refrac532 · Answer

Please help

refrac532 · Answer

@ganeshie8 @Nnesha

refrac532 · Answer

Guys please help will give medals

unklerhaukus · Answer

$$y(t) = y_0-\tfrac12gt^2$$$$x(t)=x_0+v_xt$$

refrac532 · Answer

I got about 2.69 meters...

whpalmer4 · Answer

The stone is kicked with a horizontal velocity of 3.50 m/s.  Assuming air resistance is not a factor, the time for the stone to fall can be found by dividing the distance traveled by the horizontal velocity.

The distance the stone falls is given by $$h = \frac{1}{2}gt^2$$where $g \approx 9.81 \text{ m/s}^2$
We know the time from the first step; plug that in here and find the height of the bridge.

Hint: it is not 2.69 meters.|dw:1449720532277:dw|