Tidliges posiON

Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of

WINSTONCH [101]

a₀). You know ...
       -- the object is dropped from 5 meters
   above the pavement;
         -- it falls for 0.83 second.

a₁). Without being told, you assume ...
   -- there is no air anyplace where the marshmallow travels,
   so it free-falls, with no air resistance;
   -- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .

b). You need to find how much LESS than 5 meters
   the marshmallow falls in 0.83 second.

c). You can use whatever equations you like.
   I'm going to use the equation for the distance an object falls in
   ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d). To see how this all goes together for the solution, keep reading:

The distance that an object falls in ' T ' seconds
when it's dropped from rest is

   (1/2 G) x (T²) .

On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall

   (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

   (5.00 - 3.38) = 1.62 meters

above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

7 0

3 years ago

A girl throws a marshmallow that lands in her friend’s mouth 2 m away. The girl threw the marshmallow at an angle of 30 degrees.

natima [27]

She threw the marshmallow at a speed of around 4.76 m/s.The formula for the horizontal range gives the velocity.

<h3>What is projectile motion?</h3>

The motion of an item hurled or projected into the air, subject only to gravity's acceleration, is known as projectile motion.

The item is known as a projectile, and the course it takes is known as a trajectory. Falling object motion is a simple one-dimensional kind of projectile motion with no horizontal movement.

Given data;

The marshmallow was thrown at a distance of 2 meters

Range,R = 3 m

Initial velocity,u

The angle at which the marshmallow was thrown,θ = 30°

The acceleration due to gravity,g = 9.81 m/s²

The projectile's motion is divided into two parts: horizontal and vertical motion.

$\rm R = \frac{u^2sin2 \theta }{g} \\\\ \rm R = \frac{u^2sin2 \times (30) }{9.81} \\\\ u^2=\frac{2 \times 9.81 }{sin 60^0} \\\\ u^2=22.66 \\\\ u=4.76 \ m/sec$

Hence, she throws the marshmallow at a speed of 4.76 m/sec.

To learn more about the projectile motion refer to the link;

brainly.com/question/11049671

#SPJ1

4 0

2 years ago

A comet fragment of mass 1.96 × 1013 kg is moving at 6.50 × 104 m/s when it crashes into Callisto, a moon of Jupiter. The mass o

vredina [299]

Answer:

Recoil speed, $1.17\times 10^{-5}\ m/s$

Explanation:

Given that,

Mass of the comet fragment, $m_1=1.96\times 10^{13}\ kg$

Speed of the comet fragment, $v_1=6.5\times 10^4\ m/s$

Mass of Callisto, $m_2=1.08\times 10^{23}\ kg$

The collision is completely inelastic. Assuming for this calculation that Callisto's initial momentum is zero. So,

$m_1v_1=(m_2+m_2)V$

V is recoil speed of Callisto immediately after the collision.

$V=\dfrac{m_1v_1}{(m_2+m_2)}\\\\V=\dfrac{1.96\times 10^{13}\times 6.5\times 10^4}{(1.96\times 10^{13}+1.08\times 10^{23})}\\\\V=1.17\times 10^{-5}\ m/s$