How to solve projectile motion problems with angles.

A cyclist is riding his bike up a mountain trail. When he starts up the trail, he is going 8 m/s. As the trail gets steeper, he

shusha [124]

Answer:

a) a = - 0.0833 m / s², b) t = 4.4 s

Explanation:

a) this is a kinematics exercise where the acceleration is along the inclined plane

v = v₀ - a t

a = v₀ - v / t

a = 3 - 8/60

a = - 0.0833 m / s²

b) in this case the final velocity is zero

v = v₀ - a t

0 = v₀ - at

t = v₀ / a

t = 28 / 6.4

t = 4.375 s

t = 4.4 s

3 0

3 years ago

A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum

Gekata [30.6K]

Answer:

The final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , $m_1= 0.473 kg$

mass of puck 2 , $m_2= 0.819 kg$

initial speed of puck 1 , $u_1=2.76\frac{m}{s}$

initial speed of puck 2 , $u_2=0.00\frac{m}{s}$

Final speed of puck 1 and puck 2 be $v_1\, and\, v_2$ respectively

Apply conservation of linear momentum

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

=> $0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2$

=> $1.594=0.5775\times v_1+ v_2$ -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

$u_2-u_1=v_1-v_2$

=> $0-2.76=v_1-v_2$ ------(B)

From equation (A) and (B)

$v_1=-0.739\frac{m}{s}$

and $v_2=2.02\frac{m}{s}$

Thus the final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

3 0

3 years ago

If the weight of a submerged object is equal to the buoyant force, what net force acts upon the object?

Y_Kistochka [10]

Answer:

A

Explanation:

The weight is acting downwards where as the buoyant force acting upwards (opposite) direction with equal amount of force. so the opposite forces cancel out each other (because of the force amount being equal) and no net force is acting on the object.

Hope i have helped you

Thanks.

7 0

3 years ago

. A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00m above the ground. At t

dedylja [7]

Answer:

Explanation:

Height of building

H = 6m

Horizontal speed of first balloon

U1x = 2m/s

Second ballot is thrown straight downward at a speed of

U2y = 2m/s

Time each gallon hits the ground

Balloon 1.

Using equation of free fall

H = Uoy•t + ½gt²

Uox = 0 since the body does not have vertical component of velocity

6 = ½ × 9.8t²

6 = 4.9t²

t² = 6 / 4.9

t² = 1.224

t = √1.224

t = 1.11 seconds

For second balloon

H = Uoy•t + ½gt²

6 = 2t + ½ × 9.8t²

6 = 2t + 4.9t²

4.9t² + 2t —6 = 0

Using formula method to solve the quadratic equation

Check attachment

From the solution we see that,

t = 0.9211 and t = -1.329

We will discard the negative value of time since time can't be negative here

So the second balloon get to the ground after t ≈ 0.92 seconds

Conclusion

The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.

4 0

3 years ago

Read 2 more answers

One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious spa

Ludmilka [50]

Answer:

ΔP.E = 6.48 x 10⁸ J

Explanation:

First we need to calculate the acceleration due to gravity on the surface of moon:

g = GM/R²

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of moon = 7.36 x 10²² kg

R = Radius of Moon = 1740 km = 1.74 x 10⁶ m

Therefore,

g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²

g = 2.82 m/s²

now the change in gravitational potential energy of rocket is calculated by:

ΔP.E = mgΔh

where,

ΔP.E = Change in Gravitational Potential Energy = ?

m = mass of rocket = 1090 kg

Δh = altitude = 211 km = 2.11 x 10⁵ m

Therefore,

ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)

7 0

4 years ago