Question

Practice Exercises Name: : 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? b) If the stone had been kicked harder, how would this affect the time it would take to fall

Answer 1

Answer:

a) The answer is 11,7m

b) The time it takes to fall will be shorter

Explanation:

We will use the next semi-parabolic movement equations

$H=Hi+Viy*t+1/2*g*t$

$X=Xi+Vx*t$

Where g(gravity acceleration)=9,81m/s^2

Also Xi, Hi and Viy are zero, as the stones Billy-Jones is kicking stay still before he moves them, so we take that point as the reference point

The first we must do is to find how much time the stones take to fall, this way:

$t=(5.40m)/(3.50m/s)$

Then t=1,54s

After that we need to replace t to find H, this way

$H=(1/2)*(9,81m/s^2)*(1,54s)^2$

Then H=11,7m

b) The stones will fall faster as the stones will be kicked harder, it will cause the stones move faster, it means, more horizontal velocity. In order to see it better we could assume the actual velocity is two times more than it is, so it will give us half of the time, this way:

$t=(5,40m)/(2*3,50m/s)$

Then, t=0,77