Question

Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 27.0°below the horizontal. If it strikes the ground 59.0 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.(b) the inital.(c) the speed and angle of the velocity vector with respect to the horizontal at impact.

Answer 1

a) The time of flight is 3.78 s

b) The initial speed is 17.0 m/s

c) The speed at impact is 46.4 m/s at $70.3^{\circ}$ below the horizontal

Explanation:

The picture of the previous problem (and some data) is missing: find it in attachment.

a)

The motion of the ball is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

We start by considering the vertical motion, to find the time of flight of the ball. We do it by using the following suvat equation: for the y-displacement:

$y=u_y t+\frac{1}{2}at^2$

where we have:

y = -45.0 m is the vertical displacement of the ball (the height of the building)

$u_y=u sin \theta$ is the initial vertical velocity, with u being the initial velocity (unknown) and $\theta=-27.0^{\circ}$ the angle of projection

t is the time of the fall

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Along the x-direction, the equation of motion is instead

$x=(u cos \theta)t$

where $ucos \theta$ is the horizontal component of the velocity. Rewriting this equation as

$t=\frac{x}{ucos \theta}$

And substituting into the previous equation, we get

$y=xtan \theta + \frac{1}{2}gt^2$

And using the fact that the horizontal range is

x = 59.0 m

And solving for t, we find the time of flight:

$t=\sqrt{\frac{y-x tan \theta}{g}}=\sqrt{\frac{-45-(59.0)(tan(-27^{\circ}))}{-9.8}}=3.78 s$

b)

We can now find the initial speed, u, by using the equation of motion along the x-direction

$x=u cos \theta t$

where we know that:

x = 59.0 m is the horizontal range

$\theta=-23^{\circ}$ is the angle of projection

$t=3.78 s$ is the time of flight

Solving for u, we find the initial speed:

$u=\frac{x}{cos \theta t}=\frac{59.0}{(cos (-23^{\circ}))(3.78)}=17.0 m/s$

c)

First of all, we notice that the horizontal component of the velocity remains constant during the motion, and it is equal to

$v_x = u cos \theta = (17.0)(cos (-23^{\circ})=15.6 m/s$

The vertical velocity instead changes according to the equation

$v_y = u sin \theta + gt$

Substituting all the values and t = 3.78 s, the time of flight, we find the vertical velocity at the time of impact:

$v_y = (17.0)(sin (-23^{\circ}))+(-9.8)(3.78)=-43.7 m/s$

Where the negative sign means it is downward.

Therefore, the speed at impact is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(15.6)^2+(-43.7)^2}=46.4 m/s$

while the direction is given by

$\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-43.7}{15.6})=-70.3^{\circ}$

So, $70.3^{\circ}$ below the horizontal.

Learn more about projectile motion:

brainly.com/question/8751410

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