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
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.
5 answers
typo
sqrt (22.2^2 + 16^2)
sqrt (22.2^2 + 16^2)
and
tan angle = 22.2/16
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).