Answer:
h = 204.1 m
Explanation:
Given that,
Time taken to reach the ground, t = 7 s
We need to find the height of the cliff. We can find it using second equation of kinematics as follows :
Here,
u = 0 (at rest initially)
a = g
So,
So, the cliff's height is 204.1 m.
Answer:
Explanation:
circulatory and digestive. ... An immune response is primarily due to the body's white blood cells recognizing.
D. colour blindness
"red-sensitive" just means that the nerves are able to see red light
We can see that the range from the calculation is 310 m.
<h3>What is the range of a projectile?</h3>
When we throw something from a height and it lands at the same eight, we call it a projectile motion. A projectile must always be shot at an angle. The range of the projectile is obtained by the relation; R = u^2sin2θ/g
Thus;
g = acceleration due to gravity
θ = acute angle
u = velocity
R = (60)^2 sin2(30)/10
R = 310 m
Learn more about projectile:brainly.com/question/11422992
#SPJ1
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.