Answer:
4.9 meters
Step-by-step explanation:
We know that an 11 lb bowling bowl was dropped.
So we can say that the initial speed of the bowling ball is zero ,
the time taken by the bag to free fall is given as t = 1.0 s; and
the acceleration of free fall is given as a = 9.8 m/s
.
So we will use the following equation to find the distance after falling fo 1 sec:
![d=v_i*t+(1)/(2) at^2](https://img.qammunity.org/2019/formulas/physics/high-school/gh8kp27fb0xk8xvsh9dghpouwq0uep1uqa.png)
Substituting the given values to get:
![d=0+(1)/(2) * 9.8 * 1^2](https://img.qammunity.org/2019/formulas/physics/high-school/s1wj24mn2gjocy1l0ys4meepjyfjcn3w36.png)
![d=4.9](https://img.qammunity.org/2019/formulas/physics/high-school/zrp00m1hjdgi54weuup2y8iehv4w0bozcx.png)
Therefore, the bowling ball will cover a distance of 4.9 meters when released from the top of the building.