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The final velocity of the rock before it touches the ground is 28 m/s.
Answer:
Explanation:
As the rock is thrown down, this means the acceleration due to gravity will be exerting on the rock. So the rock will be exhibiting a free fall motion. Thus, the acceleration of the rock will be equal to the magnitude of acceleration due to gravity. Then using the third equation of motion, we can determine the final velocity of the rock provided the values for initial velocity, displacement and acceleration is given in the problem itself.
So the acceleration is equal to 9.8 m/s² due to its free fall motion and displacement will be equal to the height of the tower which is given as 30 m. And the initial speed of the rock is stated as 14 m/s. The initial speed is represented as u, final speed is represented as v, displacement is represented as s and acceleration is represented as a.
Then, 2 × 9.8 × 30 = v²-(14)²
v²=784
v= 28 m/s
So the final velocity of the rock before it touches the ground is 28 m/s.
Answer:
joule
Explanation:
Let m = mass of the car and v1 = initial velocity and v2 = final velocity
Given.
Initial velocity = 100 km/h
final velocity = 50 km/h
What is work done in the car to slow it from 100km/h to 50km/h?
The work done in the car to slow it from v1 to v2.
w=Δk
joule.
Therefore, the work done is joule
Momentum = (mass) x (speed)
Change in momentum = (force) x (time)
The initial momentum is (mass) x (speed) = 2500x 25 = 62,500 kg-m/s.
Since you want to <u>stop</u> the vehicle, that number is also the required <em>change</em>
in momentum ... you want the vehicle to wind up with zero momentum.
62,500 = (force) x (time) = 20 x force
Divide each side by 20 :
force = 62,500 / 20 = <em>3,125 newtons </em>
Answer:
2.57 seconds
Explanation:
The motion of the ball on the two axis is;
x(t) = Vo Cos θt
y(t) = h + Vo sin θt - 1/2gt²
Where; h is the initial height from which the ball was thrown.
Vo is the initial speed of the ball, 22 m/s , θ is the angle, 35° and g is the gravitational acceleration, 9.81 m/s²
We want to find the time t at which y(t) = h
Therefore;
y(t) = h + Vo sin θt - 1/2gt²
Whose solutions are, t = 0, at the beginning of the motion, and
t = 2 Vo sinθ/g
= (2 × 22 × sin 35°)/9.81
= 2.57 seconds
