A ball is thrown vertically upward with speed 65m/s . the distance covered by the in 7th sec is?
1 Answer
Explanation:
We're asked to find the total distance traveled of a projectile after
We can do this by adding together two distances:
-
The distance from ground level (or from wherever it was thrown) to its maximum height.
-
The distance from its maximum height to its height at
#t = 7# #"s"# (on the way back down)
(1)
To find its maximum height, we can use the equation
where
-
#v_y# is the velocity at its maximum height, which is#0# -
#v_(0y)# is the initial velocity (#65# #"m/s"# ) -
#a_y# is the constant acceleration, which is#-g# , equal to#-9.8# #"m/s"^2# -
#Deltay# is its height (in#"m"# )
Plugging in known values, we have
It thus traveled a distance
(2)
Now, let's find its position at
To do this, we can use the equation
Plugging in values, we have
Thus, it traveled a downward distance of
Total distance
The total distance it travels is the distance traveled upward plus the distance downward:
I'll leave it to no decimal places, even though it's practically the same as the upward distance.