Answer:
The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.
Step-by-step explanation:
Let us take the point of projection of the ball as origin of the coordinate system, the upward direction as positive and down direction as negative.
Initial velocity u with which the ball is projected upwards = + 120 ft/s
Uniform acceleration a acting on the ball is to acceleration due to gravity = - 32 ft/s²
The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.
Using the formula:
v² - u² = 2 a h,
where
u = initial velocity of the ball = +120 ft/s
v = final velocity of the ball at the highest point = 0 ft/s
a = uniform acceleration acting on the ball = -32 ft/s²
h = height attained
Substituting the values we get;
0² - 120² = 2 × (- 32) h
=> h = 120²/2 × 32 = 225 feet
The height of the ball from the ground at its highest point = 225 feet + 12 feet = 237 feet.
The outlier, 78, is removed from the data set shown. 23, 23, 25, 27, 34, 34, 35, 41, 45, 45, 78 How does this affect the range?
KATRIN_1 [288]
Hello.
The range with 78 is 55
without 78 is 22
So the range goes down
Have a nice day
Answer:
the answer would be x=10
Step-by-step explanation:
cuz you do a t-chart the put x/-3 then + 10 then - witch is
x=10
