All categories

2 years ago

(i) A stone is thrown vertically upwards with an initial velocity of 60ms. Find

Physics

2 answers:

Alex777 [14]2 years ago

7 0

Answer:

i) 180m ii) 6sec ii)180m

Explanation:

i) Hmax = U²/2g

= 60²/20

= 3600/20= 180m

ii). Tmax= U/g

= 60/10= 6sec

iii). X= ut²+ 1/2gt²

since U is 0,then ut=0

X=0+ 1/2×10×6²

X=5×36= 180m

hope its clear??

elixir [45]2 years ago

5 0

Answer:

plzzzzzzz follow me back

You might be interested in

A motorcycle of mass 100 kilograms slowly rolls off the edge of a cliff and falls for three seconds before reaching the bottom o

Jet001 [13]

Answer:

C) 3,000 kg m/s

Explanation:

We can consider the horizontal velocity of the motorcycle to be zero, since it rolls off the edge of the cliff very slowly. So, we only need to find the vertical velocity at the time of the impact with the ground.

The vertical velocity of the motorcycle at time t is given by (free-fall motion):

$v(t)=v_0 -gt$

where

$v_0=0$ is the initial vertical velocity (zero, since the motorcycle is not moving)

g = 9.8 m/s^2 is the acceleration due to gravity

t is the time

Since the motorcycle hits the ground after t = 3 seconds, we have

$v(3 s)=0-(9.8 m/s^2)(3 s)=-29.4 m/s$

And since we know its mass, m=100 kg, we can find its momentum:

$p=mv=(100 kg)(-29.4 m/s)=-2940 kg m/s \cdot -3000 kg m/s$

and the negative sign simply means downward direction.

8 0

2 years ago

The half-life of iodine-131 is 8 days. Chan Hee examines two samples of iodine-131. Sample 1: 10 g Sample 2: 20 g What can Chan

Illusion [34]

Chan Hee can expect to observe about the samples after 16 days is that three-fourths of each sample will have undergone radioactive decay. The best answer is letter B.

4 0

3 years ago

A 100 kg bungee jumper leaps from a bridge. The bungee cord has an un-streched equilibrium length of 10 m, and a spring constant

nika2105 [10]

Answer:

11.78meters

Explanation:

Given data

Mass m = 100kg

Length of cord= 10m

Spring constant k= 35N/m

At the greatest vertical distance, the spring potential energy is equal to the gravitational potential energy

That is

Us=Ug

Us= 1/2kx^2

Ug= mgh

1/2kx^2= mgh

0.5*35*10^2= 100*9.81*h

0.5*35*100=981h

1750=981h

h= 1750/981

h= 1.78

Hence the bungee jumper will reach 1.78+10= 11.78meters below the surface of the bridge

6 0

2 years ago

Read 2 more answers

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

2 years ago

Read 2 more answers

A ball is thrown forward at 5 m/s how would the path of the ball differ on earth than on the moon

Levart [38]

In this problem, we are given that a ball is thrown upward and that it has an initial velocity of 5 m/s. Its path both on Earth and on the moon will be trajectory in nature. This is because while its horizontal velocity is maintained, its vertical velocity increases.

6 0

2 years ago

Read 2 more answers