Answer:
magnitude = 304.14 km/h
direction: 
West of North
Explanation:
The final plane's vector velocity will be the result of the vector addition of one pointing North of length 300 km/h, another one pointing West of length 50 km/h.
To find the magnitude of the final velocity vector (speed) we need to apply the Pythagorean theorem in a right angle triangle with sides: 300 and 50, and find its hypotenuse:
km/h
The actual direction of the plane is calculated using trigonometry, in particular with the arctan function, since the tangent of the angle can be written as:
So the resultant velocity vector of the plane has magnitude = 304.14 km/h,
and it points West of the North direction.
Answer:
t = 4.08 s
R = 40.8 m
Explanation:
The question is asking us to solve for the time of flight and the range of the rock.
Let's start by finding the total time it takes for the rock to land on the ground. We can use this constant acceleration kinematic equation to solve for the displacement in the y-direction:
We have these known variables:
- (v_0)_y = 0 m/s
- a_y = -9.8 m/s²
- Δx_y = -20 m
And we are trying to solve for t (time). Therefore, we can plug these values into the equation and solve for t.
- -20 = 0t + 1/2(-9.8)t²
- -20 = 1/2(-9.8)t²
- -20 = -4.9t²
- t = 4.08 sec
The time it takes for the rock to reach the ground is 4.08 seconds.
Now we can use this time in order to solve for the displacement in the x-direction. We will be using the same equation, but this time it will be in terms of the x-direction.
List out known variables:
- v_0 = 10 m/s
- t = 4.08 s
- a_x = 0 m/s
We are trying to solve for:
By using the same equation, we can plug these known values into it and solve for Δx.
- Δx = 10 * 4.08 + 1/2(0)(4.08)²
- Δx = 10 * 4.08
- Δx = 40.8 m
The rock lands 40.8 m from the base of the cliff.
Answer:
a= -0.86 m/s²
The negative sign shows that ball down the ground or moving down
Explanation:
Vf² - Vo² = 2gS
where
Vf = velocity of clay as it hits the ground
Vo = initial velocity of clay = 0
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
S = distance travelled by clay = 15 m
Substituting appropriate values,
Vf² - 0 = 2(9.8)(15)
Vf = 17.15 m/sec.
Formula to use is,
V - Vf = aT
where
V = velocity of clay when it stops = 0
Vf = 17.15 m/sec (as determined above)
a = acceleration
T = 20 ms
Put the values to find acceleration
a=(V-Vf)/T
a=(0-17.15)/20
a= -0.86 m/s²
The negative sign shows that ball down the ground
