Respuesta :
a)
The final speed v_f can be calculated from the formula for the acceleration:
[tex]v_f=v_0+gt[/tex]Replace v_0=30.0ft/s, g=32.2ft/s^2 and t=3.50s to find the final speed of the rock:
[tex]v_f=30.0\frac{ft}{s}+(32.2\frac{ft}{s^2})(3.50s)=142.7\frac{ft}{s}[/tex]Therefore, the speed of the rock as it hits the water is approximately 143 ft/s.
b)
The distance d traveled by an object during a time interval t is given by the average speed as follows:
[tex]d=\bar{v}\cdot t[/tex]On the other hand, an object in free fall accelerates uniformly. The average speed of an object under uniformly accelerated motion is:
[tex]\bar{v}=\frac{v_f+v_0_{}}{2}[/tex]Then, for uniformly accelerated motion, the distance traveled by an object that changes its velocity from v_0 to v_f during a time interval t, is:
[tex]d=\frac{v_f+v_0}{2}\cdot t[/tex]Notice that the initial speed of the rock is known, as well as the time interval t during which the rock falls. The final speed was calculated in part (a). Then, the distance that the rock travels during the time interval of 3.50s is:
[tex]d=\frac{142.7\frac{ft}{s}+30.0\frac{ft}{s}}{2}\times3.50s=302.225ft[/tex]Therefore, the height of the bridge above the water is approximately 302 feet.
