Given: Weight W = 100 lbs convert to Kg = 45.36 Kg
Height h = 8 Ft convert to meter = 2.44 m
Time t =10 s
Required: Power output
Formula: P = W/t Work = Force x distance or Work = mgh
P = mgh/t
P = (45.36 Kg)(9.8 m/s²)(2.44 m)/10 s
P = 108.46 J/s
P = 0.15 Horsepower
Answer:
112 m/s², 79.1°
Explanation:
In the x direction, given:
x₀ = 0 m
x = 19,500 cos 32.0° m
v₀ = 1810 cos 20.0° m/s
t = 9.20 s
Find: a
x = x₀ + v₀ t + ½ at²
19,500 cos 32.0° = 0 + (1810 cos 20.0°) (9.20) + ½ a (9.20)²
a = 21.01 m/s²
In the y direction, given:
y₀ = 0 m
y = 19,500 sin 32.0° m
v₀ = 1810 sin 20.0° m/s
t = 9.20 s
Find: a
y = y₀ + v₀ t + ½ at²
19,500 sin 32.0° = 0 + (1810 sin 20.0°) (9.20) + ½ a (9.20)²
a = 109.6 m/s²
The magnitude of the acceleration is:
a² = ax² + ay²
a² = (21.01)² + (109.6)²
a = 112 m/s²
And the direction is:
θ = atan(ay / ax)
θ = atan(109.6 / 21.01)
θ = 79.1°
Answer:
below the horizontal.
Explanation:
This is a projectile motion problem. So we are going to use uniform motion and free-fall formulas.
Since what we want is the angle of the diver when hitting the water, we are going to search for the components of the final velocity and with them compute the angle.
For the free-fall part we know:
- ,
- ,
- (remember that the initial velocity in the vertical component is zero because at the begining he only has horizontal velocity).
For the uniform motion part:
Notice that we already have the final velocity in the x coordinate (uniform motion means that the velocity is constant).
For the final velocity in y coordinate we are going to use:
,
since we get
,
,
,
.
Now we can find the angle using the tangent function and the components of the final velocity. Remember that they are related as follow:
,
where is the angle below the horizontal (the angle we are searching for).
We have that
.
First of all the project motion is in the form of a parabolic path and it is a common practice that a paper airplane always follow a parabolic path, whenever it is thrown in the air, and gravity is acting upon the plane, like any other projectile.