A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 16.0 m/s. The cliff is h = 25.0 m above a flat horizontal beach, as shown in the figure below.

How long after being released does the stone strike the beach below the cliff?
s

With what speed and angle of impact does the stone land?
m/s
° below the horizontal

for the first part i got 2.26 seconds,
but for the: WHAT SPEED AND ANGLE OF IMPACT DOES THE STONE LAND?
IM NOT GETTING THE RIGHT ANSWER FOR THE SPEED WITH THE CALCULATION IM GETTING -22.2M/S WHICH IS WRONG. AND SO IS THE ANGLE.

25 = 4.9 t^2

t = 2.26 seconds, yes

v = Vi - 9.81 t
v = -9.81(2.26) = -22.2
BUT WHAT ABOUT u ? 16 m/s horizontal
speed = sqrt (22.2^2+16)^2

tan angle below horizontal = 16/22.2

typo

sqrt (22.2^2 + 16^2)

and

tan angle = 22.2/16

thank you so much DAMON i really appreciate you help i was stuck on this problem for houra. Thank you again. and what time do you tutor and what days.

You are welcome. I am sorry; I have no schedule (retired and doing a lot of other stuff).

To determine the time it takes for the stone to strike the beach below the cliff, we can use the vertical motion equation:

h = (1/2) * g * t^2

Where h is the height of the cliff (25.0 m) and g is the acceleration due to gravity (9.8 m/s^2). Solving for time, we get:

t = sqrt(2h/g)

Substituting the given values, we have:

t = sqrt(2 * 25.0 / 9.8) = 2.26 seconds

For the second part, to find the speed and angle of impact, we can use the horizontal motion equation:

s = v * t

where s is the horizontal distance covered by the stone, v is the horizontal velocity, and t is the time (which we found to be 2.26 seconds).

Given that the stone is thrown horizontally, the initial vertical velocity (v_y) is zero, but the initial horizontal velocity (v_x) is equal to the initial speed of the stone (16.0 m/s).

Using the equation v_x = s / t, we can solve for s:

s = v_x * t = 16.0 * 2.26 = 36.2 m

So the stone covers a horizontal distance of 36.2 meters.

To find the speed of impact, we can use the Pythagorean theorem:

v_impact = sqrt(v_x^2 + v_y^2)

Since v_y is zero, this simplifies to:

v_impact = v_x = 16.0 m/s

So the stone lands with a speed of 16.0 m/s.

To determine the angle of impact below the horizontal, we can use the equation:

tan(theta) = v_y / v_x

With v_y = 0 and v_x = 16.0, we find:

tan(theta) = 0 / 16.0 = 0

Taking the inverse tangent of both sides, we find:

theta = tan^(-1)(0) = 0

Therefore, the stone lands at an angle of 0° below the horizontal.