A person throws a physics book off of a cliff at an initial speed of 3 m/s at an angle of 20 degrees. The book is in

the air for a total of 2.5 seconds.
a) How high is the cliff? b) How far away from the base of the cliff does the book land?
c) What is the magnitude and direction of the book's velocity when it lands?
d) Discuss whether or not your answers to parts a) through c) are reasonable. Give specific reasons why you
think they are or are not reasonable.

break the initial velocity into vertical and horizontal components.

Vertical:
hf=0=hi + Viv*t-1/2 g t^2 solve for hi
final vertical veloicty: Viv-g t
horizontal: Vf=Vih

magnitude of velocity:
sqrt(Vfh^2+ Vfv^2)

(a) the motion of the book is projectile, and uniformly accelerated motion. thus we use the formula,

h = ho + vo*t - (1/2)*g*t^2
where
h = final height
ho = initial height
vo,y = initial vertical velocity = vo*sin(theta)
t = time
g = acceleration due to gravity (9.8 m/s^2)
making the ground as the reference or the origin (thus at this point, the height = h = 0), we solve now for ho:
0 = ho + vo,y*t - (1/2)*g*t^2
0 = ho + 3*sin(20)*2.5 - (1/2)(9.8)(2.5)^2
0 = ho - 28.06
ho = 28.06 m

(b) since horizontal velocity, v,x is always constant, initial horizontal velocity = final horizontal velocity, or
vo,x = v,x
but vo,x = vo*cos(theta) , thus
v,x = 3*cos(20)
v,x = 2.82 m/s
we can use this to get the range (horizontal distance):
R = v,x * t
R = 2.82*2.5
R = 7.05 m

(c) note that we already solved for the final horizontal velocity, v,x = 2.82 m/s (positive/to the right)
we solve now for the final vertical velocity using
vf,y = vo,y - g*t
substituting,
vf,y = 3*sin(20) - 9.8*2.5
vf,y = -23.5 m/s (negative/downward direction)
getting direction,
arctan(23.5/2.82) = 83.1 degrees

hope this helps~ :)

To find the answers to these questions, we can break down the motion of the book into horizontal and vertical components. Let's go through each part step by step:

a) How high is the cliff?

To find the height of the cliff, we need to determine the time it takes for the book to reach its highest point. We can do this by finding the time it takes for the vertical component of the book's velocity to become zero.

Given that the book is in the air for a total of 2.5 seconds, we can divide this time in half to find the time it takes for the book to reach its highest point, as the book will have equal upward and downward travel times.

Therefore, the time it takes for the book to reach its highest point is 2.5 seconds divided by 2, which is 1.25 seconds.

Using this time, we can use the kinematic equation for vertical displacement:

\[y = v_iy*t + 0.5*a_y*t^2 \]

Since the initial vertical velocity \(v_iy\) is given by \(v_i*sin(\theta)\), where \(\theta\) is the angle of the initial velocity (20 degrees) and \(a_y\) is the acceleration due to gravity (-9.8 m/s^2), we can find the vertical displacement.

\[y = (3 m/s * sin(20°)) * 1.25 s + 0.5*(-9.8 m/s^2)*(1.25 s)^2 \]

Calculating this expression will give us the height of the cliff.

b) How far away from the base of the cliff does the book land?

To determine the horizontal distance traveled by the book, we need to find the time it takes for the book to reach the ground. We can use the total time the book is in the air, which is given as 2.5 seconds, to calculate this.

Since we know the horizontal velocity \(v_ix\) is given by \(v_i*cos(\theta)\), where \(\theta\) is the angle of the initial velocity (20 degrees), we can find the horizontal distance traveled.

\[x = v_ix * t \]

Calculating this expression will give us the distance the book lands from the base of the cliff.

c) What is the magnitude and direction of the book's velocity when it lands?

To find the magnitude and direction of the book's velocity when it lands, we need to calculate the final velocity \(v_f\) in both the horizontal and vertical components.

The final vertical velocity \(v_fy\) is given by \(v_iy + a_y * t\), where \(v_iy\) is the initial vertical velocity (given by \(v_i * sin(\theta)\)), \(a_y\) is the acceleration due to gravity (-9.8 m/s^2), and \(t\) is the total time the book is in the air (2.5 seconds).

The final horizontal velocity \(v_fx\) remains constant throughout the entire motion since there is no horizontal acceleration. Therefore, \(v_fx = v_ix\) (given by \(v_i * cos(\theta)\)).

The magnitude of the book's velocity when it lands can be calculated as \(v = \sqrt{v_fx^2 + v_fy^2}\), and the direction can be determined by finding the angle \(\theta\) using trigonometry with \(v_fx\) and \(v_fy\).

d) Discuss whether or not your answers to parts a) through c) are reasonable. Give specific reasons why you think they are or are not reasonable.

To evaluate the reasonability of the answers, we need to consider factors such as the height and distance traveled by the book, as well as the magnitude and direction of its velocity.

For example, if the height of the cliff calculated seems extremely high or unrealistic based on the physical surroundings or context provided, it may raise doubts about the accuracy of the calculations.

Similarly, if the distance traveled by the book seems too short or too long relative to the initial velocity and the time it stays in the air, it may also be a cause for concern.

Lastly, the magnitude and direction of the book's velocity when it lands should align with our expectations based on the initial conditions and the laws of physics. If the magnitude is too high or too low, or if the direction is significantly different from the expected path, it could indicate errors in the calculations or assumptions made.

Overall, reasonability can be assessed by comparing the calculated values to our intuition and understanding of the physical principles involved, and by considering any potential limitations or uncertainties in the given data or assumptions.