Answer:
about 19.6° and 73.2°
Explanation:
The equation for ballistic motion in Cartesian coordinates for some launch angle α can be written ...
y = -4.9(x/s·sec(α))² +x·tan(α)
where s is the launch speed in meters per second.
We want y=2.44 for x=50, so this resolves to a quadratic equation in tan(α):
-13.6111·tan(α)² +50·tan(α) -16.0511 = 0
This has solutions ...
tan(α) = 0.355408 or 3.31806
The corresponding angles are ...
α = 19.5656° or 73.2282°
The elevation angle must lie between 19.6° and 73.2° for the ball to score a goal.
_____
I find it convenient to use a graphing calculator to find solutions for problems of this sort. In the attachment, we have used x as the angle in degrees, and written the function so that x-intercepts are the solutions.
Sand is the largest particle that can generally be carried by the wind.
Answer:
The chunk went as high as
2.32m above the valley floor
Explanation:
This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.
Applying the principle of energy conservation for the two ice we have based on the scenery
Momentum before impact = momentum after impact
M1U1+M2U2=(M1+M2)V
Given data
Mass of ice 1 M1= 5.20kg
Mass of ice 2 M2= 5.20kg
velocity of ice 1 before impact U1= 13.5 m/s
velocity of ice 2 before impact U2= 0m/s
Velocity of both ice after impact V=?
Inputting our data into the energy conservation formula to solve for V
5.2*13.5+5.2*0=(10.4)V
70.2+0=10.4V
V=70.2/10.4
V=6.75m/s
Therefore the common velocity of both ice is 6.75m/s
Now after impact the chunk slide up a hill to solve for the height it climbs
Let us use the equation of motion
v²=u²-2gh
The negative sign indicates that the chunk moved against gravity
And assuming g=9.81m/s
Initial velocity of the chunk u=0m/s
Substituting we have
6.75²= 0²-2*9.81*h
45.56=19.62h
h=45.56/19.62
h=2.32m
Answer:
answer is iron rod bcoz its a conductor.