Physics, 01.10.2019 20:00 JaredO

Suppose the ball is thrown from the same height as in the practice it problem at an angle of 39.0° below the horizontal. if it strikes the ground 41.8 m away, find the following. (hint: for part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.) (a) the time of flight s (b) the initial speed m/s (c) the speed and angle of the velocity vector with respect to the horizontal at impact speed m/s angle ° below the horizontal

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The data about the previous problem are missing: I found them online and attached here (see picture). The relevant missing data is just the height of the building, which is 45 meters.

(a) 1.51 s

The motion along the x-direction is uniform (constant speed), so the position along the x-direction is given by

$x(t) = (v_0 cos \theta) t$

where v_0 is the initial speed, $\theta=39.0^{\circ}$ is the angle of launching, t is the time. We can re-write the formula as

$v_0 t = \frac{x(t)}{cos \theta}$ (1)

Along the y-direction the motion is accelerated (free-fall), so the equation for the vertical position is

$y(t)=h-v_0 sin \theta t -\frac{1}{2}gt^2$ (2)

where h = 45.0 m is the height of the building and g=9.8 m/s^2 is the acceleration of gravity. Substituting (1) into (2),

$y(t)=h-x(t) tan \theta -\frac{1}{2}gt^2$

The ball strikes the ground when y(t)=0, so the equation becomes

$0=h-x(t) tan \theta -\frac{1}{2}gt^2$

And when this happens, the displacement along the x-axis is x = 41.8 m. Solving the equation for t and substituting the numbers, we find the time of flight:

$t=\sqrt{\frac{2(h-xtan \theta)}{g}}=\sqrt{\frac{2(45-41.8 tan 39^{\circ})}{9.8}}=1.51 s$

(b) 35.6 m/s

We can now easily find the initial speed by looking at the equation for the horizontal displacement:

$x(t) = (v_0 cos \theta) t$

Re-arranging it,

$v_0=\frac{x(t)}{t cos \theta}$

And substituting:

x = 41.8 m

t = 1.51 s

$\theta=39.0^{\circ}$

we find

$v_0 = \frac{41.8}{(1.51)(cos 39.0^{\circ})}=35.6 m/s$

The horizontal component of the velocity during the whole motion is constant, so it is

$v_x = v_0 cos \theta = (35.6)(cos 39.0^{\circ})=27.7 m/s$

The vertical velocity instead follows the equation

$v_y(t) = -v_0 sin \theta -gt$

where the negative signs means the direction is downward. Substituting t=1.51 s, we find the vertical velocity at the time of impact:

$v_y =-(35.6)(sin 39.0^{\circ})-(9.8)(1.51)=-37.2 m/s$

Therefore, the speed at the moment of impact is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(27.7)^2+(-37.2)^2}=46.4 m/s$

And the direction will be given by

$\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-37.2}{27.7})=-53.3^{\circ}$

Suppose the ball is thrown from the same height as in the practice it problem at an angle of 39.0° b

Answer from: Quest

the average speed is 24.57 km/hr

Answer from: Quest

Julia would be observing a. static electricity was caused by her sweater rubbing against her shirt and transferring electrons. she didn't touch anything metal, so that would cancel out b.she most likely didn't rub her sweater against her hair either, so that cancels out c.she also didn't scoot across the carpet while trying to take her sweater off, that cancels out d.so that leaves a.a. is correct.

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Suppose the ball is thrown from the same height as in the practice it problem at an angle of 39.0° b...