A ball is thrown vertically downward from the top of a 30.6-m-tall building. The ball passes the top of a window that is 10.7 m
1 answer:
Answer:
v = 19.6 m/s
Explanation:
Height of building = 30.6 m.
Height of window from the ground level= 10.7 m.
Acceleration due to gravity = 9.8
At initial condition ball at rest condition so u= 0 m/s.
Lets take when passes through the window ,velocity is v.
Here acceleration is constant so we can apply motion equation .
We know that
v= u + a t
So by putting the values
v = 0 +9.8 x 2
v = 19.6 m/s
So the velocity of ball is 19.6 m/s when passes through the window after 2 s.
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Answer:
Distance: 21 yd, displacement: 15 yd, gain in the play: 12 yd
Explanation:
The distance travelled by Sam is just the sum of the length of each part of Sam's motion, regardless of the direction. Initially, Sam run from the 3 yd line to the 15 yd line, so (15-3)=12 yd. Then, he run also 9 yd to the right. Therefore, the total distance is
d = 12 + 9 = 21 yd
The displacement instead is a vector connecting the starting point with the final point of the motion. Sam run first 12 yd straight ahead and then 9 yd to the right; these two motions are perpendicular to each other, so we can find the displacement simply by using Pythagorean's theorem:
Finally, the yards gained by Sam in the play are simply given by the distance covered along the forward-backward direction only. Since Sam only run from the 3 yd line to the 15 yd line along this direction, then the gain in this play was
d = 15 - 3 = 12 yd