Question #250941

A stonc is thrown horizontally out to sea from the

top of a vcrtical clilf and with a specd of l0 Oms'''

The ciiflis 122 5m high. Find

(l) thc timc it takes the stone to hit the rvater;

(2) the speed of the stone ald the direction in which

it is fravclling when it hits the water;

(3) the horizontal distance fromthe bottom ofthe

clilf at which the stone hits the water'

Ifthc stone:hits.a seagull,3 seconds after it is

thrown, find

(4) its velocity when it hits the seagull;

(5) how lar the seagull was from the clilf when it

was hit;

(6) the height above the water at which the seagull

was flYrng.



1
Expert's answer
2021-10-13T15:42:16-0400

(1) The time is


T=2h/g=5 s.T=\sqrt{2h/g}=5\text{ s}.

(2) The magnitude of speed is


v=vy2+vx2=2gh+vx2,v=29.8122.5+102=50 m/s. θ=arctan2ghvx=89.76°.v=\sqrt{v_y^2+v_x^2}=\sqrt{2gh+v_x^2},\\ v=\sqrt{2·9.8·122.5+10^2}=50\text{ m/s}.\\\space\\ \theta=\arctan\frac{2gh}{v_x}=89.76°.

(3) The range is


R=vxT=50 m.R=v_xT=50\text{ m}.

(5) The distance was


Rs=vxt=103=30 m.R_s=v_xt=10·3=30\text{ m}.

(4) The vertical speed and the speed of impact with the gull was


vy=gt=29.4 m/s, v=vy2+vx2=31.05 m/s.v_y=gt=29.4\text{ m/s},\\\space\\ v=\sqrt{v_y^2+v_x^2}=31.05\text{ m/s}.

6) The height was


hy=hgt22=78.4 m.h-y=h-\frac{gt^2}2=78.4\text{ m}.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS