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Furkat [3]
3 years ago
6

Jocko the clown, whose mass is 60-Kg, stands on a skateboard. A 20-Kg ball is thrown at Jocko at 3m/s, and when he catches the b

all, he and the ball move on the skateboard. How fast do Jocko and the ball move after he catches the ball?
Physics
1 answer:
Mekhanik [1.2K]3 years ago
7 0

Answer:

The speed of the Jocko and the ball move after he catches the ball is 0.75 m/s.

Explanation:

Given that,

Mass if Jocko, m = 60 kg

Mass of the ball, m' = 20 kg

Speed of the ball, v = 3 m/s

Let V is the speed of Jocko and the ball move after he catches the ball. The momentum of the system remains conserved. Using the conservation of momentum as :

m'v'=(m+m')V\\\\V=\dfrac{m'v'}{(m+m')}\\\\V=\dfrac{20\times 3}{(60+20)}\\\\V=0.75\ m/s

So, the speed of the Jocko and the ball move after he catches the ball is 0.75 m/s.

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2. A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration of 2.5 . At 30.0 s after blast
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Answer:

a)The highest point reached by the rocket is 1412 m

b)The rocket crashes after 54.7 s

Explanation:

Hi there!

The equations of height and velocity of the rocket are the following:

h = h0 + v0 · t + 1/2 · a · t² (while the engines work).

h = h0 + v0 · t + 1/2 · g · t² (when the rocket is in free fall).

v = v0 + a · t (while the engines work).

v = v0 + g · t (when the rocket is in free fall).

Where:

h = height of the rocket at a time t.

h0 = initial height of the rocket.

v0 = initial velocity.

t = time.

a = acceleration due to the engines.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the rocket at a time t.

First, let's find the velocity and height reached by the rocket until the engines fail:

h = h0 + v0 · t + 1/2 · a · t²

Let's set the origin of the frame of reference at the launching point so that h0 = 0. Since the rocket starts from rest, v0 = 0. So after 30.0 s the height of the rocket will be:

h = 1/2 · a · t²

h = 1/2 · 2.5 m/s² · (30.0 s)²

h = 1125 m

Now let's find the velocity of the rocket at t = 30.0 s:

v = v0 + a · t (v0 = 0)

v = 2.5 m/s² · 30.0 s

v = 75 m/s

After 30.0s the rocket will continue to ascend with a velocity of 75 m/s. This velocity will be gradually reduced due to the acceleration of gravity. When the velocity is zero, the rocket will start to fall. At that time, the rocket is at its maximum height. So, let's find the time at which the velocity of the rocket is zero:

v = v0 + g · t

0 = 75 m/s - 9.8 m/s² · t (v0 = 75 m/s because the rocket begins its free-fall motion with that velocity).

-75 m/s / -9.8 m/s² = t

t = 7.7 s

Now, let's find the height of the rocket 7.7 s after the engines fail:

h = h0 + v0 · t + 1/2 · g · t²

The rocket begins its free fall at a height of 1125 m and with a velocity 75 m/s, then, h0 = 1125 m and v0 = 75 m/s:

h = 1125 m + 75 m/s · 7.7 s - 1/2 · 9.8 m/s² · (7.7 s)²

h = 1412 m

The highest point reached by the rocket is 1412 m

b) Now, let's calculate how much time it takes the rocket to reach a height of zero (i.e. to crash) from a height of 1412 m.

h = h0 + v0 · t + 1/2 · g · t² (v0 = 0 because at the maximum height the velocity is zero)

0 = 1412 m - 1/2 · 9.8 m/s² · t²

-1412 m / -4.9 m/s² = t²

t = 17 s

The rocket goes up for 30.0 s with an acceleration of 2.5 m/s².

Then, it goes up for 7.7 s with an acceleration of -9.8 m/s².

Finally, the rocket falls for 17 s with an acceleration of -9.8 m/s²

The rocket crashes after (30.0 s + 7.7 s + 17 s) 54.7 s

6 0
3 years ago
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