Ask question

Ask question

All categories

alexandr402 [8]

3 years ago

7

An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b

) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

1 answer:

lys-0071 [83]3 years ago

3 0

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

S = ut + 0.5at²

Here u = 0, and a = g

S = 0.5gt²

Distance traveled during the first second ( t =1 )

S = 0.5 x 9.81 x 1² = 4.905 m

Distance traveled during the first second = 4.905 m.

b) We have equation of motion

v² = u² + 2as

Here u = 0, s= 75 m and a = g

v² = 0² + 2 x g x 75 = 150 x 9.81

v = 38.36 m/s

Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

75 = 0.5 x 9.81 x t²

t = 3.91 s

We need to find distance traveled last second

That is

S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

Distance traveled during the last second of motion before hitting the ground = 33.45 m

You might be interested in

Hoover Dam on the Colorado River is the highest dam in the United States at 221m, with a power output of 680 MW. The dam generat

Phantasy [73]

Answer:

<u> Power = 9.75 ×10^8 $\frac{kgm^2}{s^3}$ </u>

Explanation:

Power is rate of change of energy.
Here gravitational energy is transferred to kinetic energy of water at a definite rate.

For one second 650m^3 of water flows out down to 150m oh depth.

So, the energy at a height of 150m is transformed to kinetic energy.

for a second,

650m^3 of water flows down ⇒ (1000kg/m^3 × 650m^3) = 6.5×10^5kg of warer flos down.

The total gravitational potential energy stored in water is

= <u>mass of water × height× gravity</u>

= 6.5 ×10^5 × 150 × 10 = 9.75 ×10^8 $\frac{kgm^2}{s^2}$

As it is transformed in a second it is also equal to <u>Power.</u>

4 0

3 years ago

Read 2 more answers

A skater increases her speed uniformly from 2.0 meters per second to 7.0 meters per second over a distance of 12 meters. the mag

AVprozaik [17]

<span>The magnitude of her acceleration as she travels this 12 meters is 1.875m/s^2</span>

6 0

3 years ago

A truck is traveling east at 80 km/h. At an intersection 32 km ahead, a car is traveling north at 50 km/h. How long after this m

nirvana33 [79]

The time elapsed when the vehicles are closest to each other is 20 min.

The given parameters:

Speed of the truck, u = 80 km/h
Distance, d = 32 km
Speed of the car, v = 50 km/h

<h3>Principles of relative speed</h3>

The time elapsed when the cars are close to each other is calculated by applying the principles of relative speed.

$(V_r) t = d\\\\V_r^2 = 50^2 + 80^2\\\\V_r =\sqrt{50^2 + 80^2} \\\\V_r = 94.34 \ km/h$

$94.34 t = 32\\\\t = \frac{32}{94.34} \\\\t = 0.34 \ hr\\\\t \approx 20 \min$

Thus, the time elapsed when the vehicles are closest to each other is 20 min.

Learn more about relative velocity here: brainly.com/question/24430414

3 0

2 years ago

Help please! This question is driving me crazy

Varvara68 [4.7K]

Answer:

-10.8°, or 10.8° below the +x axis

Explanation:

The x component of the resultant vector is:

x = 3.14 cos(30.0°) + 2.71 cos(-60.0°)

x = 4.07

The y component of the resultant vector is:

y = 3.14 sin(30.0°) + 2.71 sin(-60.0°)

y = -0.777

Therefore, the angle between the resultant vector and the +x axis is:

θ = atan(y / x)

θ = atan(-0.777 / 4.07)

θ = -10.8°

The angle is -10.8°, or 10.8° below the +x axis.

3 0

3 years ago

A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta

larisa [96]

Answer:

Part a)

$x = 15.76 m$

Part b)

$y = 7.94 m$

Part c)

$x = 26.16 m$

Part d)

$y = 7.49 m$

Part e)

$x = 83.23 m$

Part f)

$y = -75.6 m$

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

$v_x = 19.9 cos39.9$

$v_x = 15.3 m/s$

similarly we have

$v_y = 19.9 sin39.9$

$v_y = 12.76 m/s$

Part a)

Horizontal displacement in 1.03 s

$x = v_x t$

$x = (15.3)(1.03)$

$x = 15.76 m$

Part b)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.03) - 4.9(1.03)^2$

$y = 7.94 m$

Part c)

Horizontal displacement in 1.71 s

$x = v_x t$

$x = (15.3)(1.71)$

$x = 26.16 m$

Part d)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.71) - 4.9(1.71)^2$

$y = 7.49 m$

Part e)

Horizontal displacement in 5.44 s

$x = v_x t$

$x = (15.3)(5.44)$

$x = 83.23 m$

Part f)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(5.44) - 4.9(5.44)^2$

$y = -75.6 m$

6 0

3 years ago