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:
h' = 55.3 m
Explanation:
First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:
s = vt
where,
s = horizontal distance between arrow and orange = 60 m
v = initial horizontal speed of the arrow = v₀ Cos θ
θ = launch angle = 30°
v₀ = launch speed = 35 m/s
Therefore,
60 m = (35 m/s)Cos 30° t
t = 60 m/30.31 m/s
t = 1.98 s
Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:
h = Vi t + (1/2)gt²
where,
Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°
Vi = 17.5 m/s
Therefore,
h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²
h = 34.6 m + 19.2 m
h = 53.8 m
since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:
h' = h + y
h' = 53.8 m + 1.5 m
<u>h' = 55.3 m</u>
Deposition moves that have already been eroded, and builds up sediment.
erosion is just the moving of sediment from one place to another
Answer:
The value is
Explanation:
From the question we are told that
The mass of the first car is 
the initial speed of the car is 
The mass of the second car is 
The distance move by both cars is s = 1.93 m
Generally from the law of momentum conservation

Here
because the second car is at rest
and
is the final velocity of the the two car
So
=> 
Generally from kinematic equation

here a is the deceleration
So

=> 
Generally the frictional force is equal to the force propelling the car , this can be mathematically represented as

Here F is mathematically represented as

and
So
=>
=>
Answer:
im sorry for doing this but i need points
Explanation: