Answer:
Solution given:
height [H]=25m
initial velocity [u]=8.25m/s
g=9.8m/s
now;
a. How long is the ball in flight before striking the ground?
Time of flight =?
Now
Time of flight=
substituting value
- =
- =2.26seconds
<h3>
<u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>
b. How far from the building does the ball strike the ground?
<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?
we have
Horizontal range=u*
<h3>
<u>The ball strikes 18.63m far from building</u>. </h3>
Answer:
Check the attached image
Explanation:
To solve the problem for time you will have to use the formula for time, t = d/s which means time equals distance divided by speed.
Kindly check the attached image below for the step by step explanation to the question.
If it is s-t graph , point is c
if it is v-t graph , point is e
Answer:
P₁- P₂ = 91.1 10³ Pa
Explanation:
For this exercise we will use Bernoulli's equation, where point 1 is at the bottom of the house and point 2 on the second floor
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
P1-P2 = ½ ρ (v₂² - v₁²) + ρ g (y₂-y₁)
In the exercise they give us the speeds and the height of the turbid, so we can calculate the pressure difference
For heights let's set a reference system on the ground floor of the house, so we have 5m for the second floor and an entrance at -2m
P₁-P₂ = ½ 1.0 10³ (7² - 2²) + 1.0 10³ 9.8 (5 + 2)
P₁-P₂ = 22.5 10³ + 68.6 10³
P₁- P₂ = 91.1 10³ Pa
