Please help me it is due for today i have 2 hours only left

a football is thrown upward at a 31° angle to the horizontal.the acceleration of gravity is 9.8m/s^2. To throw the ball a distan

mr Goodwill [35]

The ball's vertical position $y$ in the air at time $t$ is

$y=v_0\sin31^\circ\,t-\dfrac g2t^2$

The ball is at its original height when $y=0$ , which happens at

$v_0\sin31^\circ\,t-\dfrac g2t^2=\dfrac t2\left(2v_0\sin31^\circ-gt)=0$

$\implies t=0\text{ and }t=\dfrac{2v_0\sin31^\circ}g$

Meanwhile, the ball's horizontal position $x$ at time $t$ is

$x=v_0\cos31^circ\,t$

So when the ball reaches its original height a second time, the ball will have traveled a horizontal distance of

$x=\dfrac{2{v_0}^2\sin31^\circ\cos31^\circ}g=\dfrac{{v_0}^2\sin(2\cdot31^\circ)}g$

(which you might recognize as the formula for the range of a projectile)

To reach a distance of $x=77\,\rm m$ , the initial speed $v_0$ would be

$77\,\mathrm m=\dfrac{{v_0}^2\sin62^\circ}{9.8\,\frac{\rm m}{\mathrm s^2}}\implies v_0=29\dfrac{\rm m}{\rm s}$

7 0

3 years ago

A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or sh

zalisa [80]

Complete question:

Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. In contrast, a passenger not wearing a seat belt or using an air bag decelerates over a distance of 5mm.

(a) A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.

Answer:

The net force on the person as the air bad deploys is -6750 N backwards

Explanation:

Given;

mass of the passenger, m = 60 kg

velocity of the car at impact, u = 15 m/s

final velocity of the car after impact, v = 0

distance moved as the front of the car crumples, s = 1 m

First, calculate the acceleration of the car at impact;

v² = u² + 2as

0² = 15² + (2 x 1)a

0 = 225 + 2a

2a = -225

a = -225 / 2

a = -112.5 m/s²

The net force on the person;

F = ma

F = 60 (-112.5)

F = -6750 N backwards

Therefore, the net force on the person as the air bad deploys is -6750 N backwards

4 0

3 years ago

A 750 kg car is moving at 20.0 m/s at a height of 5.0 m above the bottom of a hill when it runs out of gas. From there, the car

S_A_V [24]

Answer:

Explanation:

Kinetic energy at the height = 1/2 m v²

= 1/2 x 750 x 20²

= 150000 J

Its potential energy = mgh

= 750 x 9.8 x 5

=36750 J

Total energy = 186750 J

Its total kinetic energy will be equal to 186750 J , according to conservation of mechanical energy

If v be its velocity at the bottom

1/2 m v² = 186750

v = √498

= 22.31 m /s

5 0

3 years ago

YALL HELP ME QUICK IN 10 MINS<br> WITH WORKING OUT

svetlana [45]

Answer:

bro I'm sorry I'm only in 7th grade

8 0

3 years ago

Read 2 more answers

Charge is distributed uniformly along a long straight wire. the electric field 2 cm from the wire is 20 n/c. the electric field

Serjik [45]

The electric field 4cm from the wire is

$É = 10nc { }^{ - 1}$

Given:

charge is distributed uniformly along a long straight wire.

electric field 2cm from the wire is 20n/c

To find:

electric field 4cm from the wire

what is electric field?

Electric field is the region around charge particle or charged body in which if another charge is placed it experiences electrostatic force.An electric field is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them.

$E ∝ \frac{1}{r}$