Explanation:
It is given that, the height of a certain tower is 862 feet i.e to reach on the ground the object should travel, s = 862 feet
The distance traveled by a freely falling object is given by :
t = 7.34 seconds
So, the object will take 7.34 seconds to fall to the ground from the top of the building. Hence, this is the required solution.
Answer:
<em>Plane Motion</em>
<em>Plane Motion One of the most common examples of motion in a plane is Projectile motion.</em>
The equation to be used here is the trajectory of a projectile as written below:
y = xtanθ +/- gx²/2v²(cosθ)²
where
y is the vertical distance
x is the horizontal distance
θ is the angle of trajectory or launch angle
g is 9.81 m/s²
v is the initial velcity
Since the angle is below horizontal, let's use the minus equation. Substituting the values:
- 0.8 m = xtan15° - (9.81 m/s²)x²/2(4.8 m/s)²(cos15°)²
Solving for x,
x = 2.549 m
However, we only take half of this distance because it was specified that the distance asked before bouncing. Hence, the horizontal distance is equal to 1.27 m.
Answer:
z = 93.2 m
Explanation:
We can appreciate that this expression is equivalent to the linear motion equation with constant acceleration
v² = v₀² + 2 a d
If we make a term-to-term comparison with the expression obtained, they are equivalent
u² = v² + 2 a z
From here we can clear the position
2 a z = u² –v²
z = (u² –v²) / 2 a
Let's calculate
For the speed to reduce the acceleration must be negative
z = (0 - 21.8²) / 2(- 2.55)
z = 93.2 m
