<h2>Answer: 10.52m</h2><h2 />
First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).
According to this, the initial velocity 
has two components, because the brick was thrown at an angle 
:
(1)
(2)
(3)
(4)
As this is a projectile motion, we have two principal equations related:
<h2>
In the x-axis:
</h2>
(5)
Where:
is the distance where the brick landed
is the time in seconds
If we already know 
and 
, we have to find the time (we will need it for the following equation):
(6)
(7)
<h2>
In the y-axis:
</h2>
(8)
Where:
is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>
is the acceleration due gravity
Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:
(9)
(10)
Multiplying by -1 each side of the equation:
>>>>This is the height of the building
8.8 miles. You divide 22.4 & 2.55.
Answer:
1. 2.5s
Explanation:
1. For time, divide Distance / speed
25m / 10
=2.5s
Answer:
C 80 m
Explanation:
Given:
v₀ = 30 m/s
a = -10 m/s²
t = 8 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (30 m/s) (8 s) + ½ (-10 m/s²) (8 s)²
Δy = -80 m
The ball lands 80 m below where it started. So the height of the cliff is 80 m.
Answer:
R = 8.01 m
Explanation:
We can solve this problem using the projectile launch equations. The jump length is the throw range
R = v₀² sin 2θ / g
in the exercise they give us the initial speed of 9.14 m / s and in the launch angle 35º
let's calculate
R = 9.14² sin (2 35) / 9.8
R = 8.01 m
this is the jump length
