Ask question

Ask question

All categories

4 months ago

9

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

eaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 8 m above the ground, and takes 1.505 s to get directly over Julie's head.
What is the speed of the ball when it leaves Sarah's hand? (the answers is not 16.67m/s)

2)How high above the ground will the ball be when it gets to Julie?(the answers is not 7.744)

1 answer:

Softa [3K]4 months ago

5 0

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle $\theta$ to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

where $y_o$ represents the initial height from ground level, $v_{oy}$ is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

$v_y=v_{oy}-gt$

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of $v_o$ .

The peak height is attained when $v_y=0$ , which allows us to compute the time to reach that height.

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum height reached is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is equal to 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Continuing with the calculations for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Substituting known values yields

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

You might be interested in

A parachutist, after opening her parachute, finds herself gently floating downward, no longer gaining speed. She feels the upwar

Sav [3153]

As the parachutist is descending at a steady rate

we can conclude that

$a = \frac{dv}{dt}$

Acceleration indicates the change in velocity

given the constant velocity in this scenario

$a = 0$

Thus, in this situation, we find the acceleration to be zero

It’s understood from Newton's second law

$F_{net} = ma$

where a is equal to 0

$F_{net} = 0$

$F_{net} = F_g - F_b$

Here, the force due to gravity

equals the force due to buoyancy

$F_g$ Hence, we can deduce

$F_b$

therefore

$F_g - F_b = 0$

as such the upward force is counteracted by the downward force.

5 0

2 months ago

While dangling a hairdryer by its cord, you observe that the cord is vertical when the hairdryer isoff and, once it is turned on

Yuliya22 [3333]

Answer:

The air exiting from the hairdryer is moving at a speed of 10 m/s.

Explanation:

The thrust generated by the hairdryer enables it to maintain an elevation angled at 5° from vertical; thus, we derive from the force diagram

$(1).\: tan (5^o) = \dfrac{F_t}{Mg}$

by substituting $M =0.420kg$ , $g = 9.8m/s^2$ into the equation and resolving for $F_t$ we find:

$F_t = Mg\:tan(5^o)$

$F_t = (0.420kg)(9.8m/s^2)\:tan(5^o)$

$\boxed{F_t = 0.3601N.}$

This thrust is linked to the speed of air ejection $v$ through the equation

$(2).\: F_t = v\dfrac{dM}{dt}$

where $dM/dt$ signifies the rate of air ejection, which is known to be

$0.06m^3/2s = 0.03m^3/s$

and since $1m^3 = 1.2kg$ ,

$0.03m^3/s \rightarrow 0.036kg/s$

$\dfrac{dM}{dt} = 0.036kg/s,$

by inserting these values into equation (2), we obtain the value of $F_t$ as:

$0.3601N = 0.036v$

resulting in

$v= \dfrac{0.3601}{0.036}$

$\boxed{v =10m/s.}$

which indicates the air velocity discharged from the hairdryer.

6 0

4 months ago

Tiana jogs 1.5 km along a straight path and then turns and jogs 2.4 km in the opposite direction. She then turns back and jogs 0

inna [3103]

Answer:

Distance: 4.6 km Displacement= -0.2 km

Explanation:

The overall distance covered: 1.5 + 2.4 + 0.7 = 4.6 km

Displacement calculation: 1.5 - 2.4 + 0.7 = -0.2 km

The displacement could also simply be stated as 0.2 km depending on whether negative value is preferred.

7 0

3 months ago