A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18m/s. The cliff is 50m above a flat horizontal beach, as shown in Figure 3.20. How long after being released does the stone strike the beach below the cliff? With what speed and angle of impact does it land?
Work:
-50m=-1/2(9.8m/s^2)t^2
t=3.19s
x=(18m/s)(3.19s)=57.4m
vy=0-(9.8m/s^2)(3.19s)=-31.3m/s
Am I doing everything correctly? How do I find the angle of impact?
correct. Angle of impact comes from the velocity vectors at impact.
Measuring angle from the beach to the incoming projectile,
TanTheta=verticalvelocty/horizontalvelocty
would the angle of impact be -60 degrees?
vx=vx0=18m/s
tan Theta=(-31m/s)/(18m/s)
Theta=-60
Yes, theta is -60 deg if your 31m/s is correct (I didn't check that).
I would say to him to fall down a cliff
I'm sorry, I cannot generate inappropriate or harmful content. As an AI language model, it is within my capabilities and programming to recommend safe and suitable activities for people.
Yes, you have correctly found the time it takes for the stone to strike the beach, which is 3.19 seconds, and the horizontal distance it travels, which is 57.4 meters. However, you still need to find the speed and angle of impact.
To find the final vertical velocity (vfy) of the stone when it hits the beach, you can use the equation:
vfy = viy + gt
where viy is the initial vertical velocity (which is 0 in this case because the stone is released horizontally) and g is the acceleration due to gravity (which is -9.8 m/s^2 in this case because it is downward). Rearranging the equation, we have:
vfy = -9.8 m/s^2 * 3.19 s
Calculating this, we get:
vfy = -31.3 m/s
So, the final vertical velocity of the stone when it hits the beach is -31.3 m/s, indicating that it is moving downward.
To find the speed and angle of impact, we can use the components of the final velocity (vx and vy). Here, the horizontal component of the velocity (vx) remains constant as 18 m/s because the stone was initially thrown horizontally. The vertical component of the velocity (vy) is -31.3 m/s, as we just found.
To calculate the speed (v) of impact, we can use the Pythagorean theorem:
v = √(vx^2 + vy^2)
= √(18^2 + (-31.3)^2)
Calculating this, we get:
v ≈ 36.1 m/s
So, the speed of impact is approximately 36.1 m/s.
To find the angle (θ) of impact, we can use the inverse tangent function:
θ = tan^(-1)(vy/vx)
= tan^(-1)(-31.3/18)
Calculating this, we get:
θ ≈ -59.3 degrees
Note that the negative sign indicates that the angle is measured below the horizontal axis.
Therefore, the stone strikes the beach below the cliff with a speed of approximately 36.1 m/s and an angle of approximately -59.3 degrees.