physics - please make it visible QUESTION 3An arrow is shot from a height h toward a clif.It is shot with a velocity Vat an angle of 8 above the horizontal.It lands on the top edge of the cliff time t later.What is the height of the cliff?O h +t V sin© + g t2/2,where g = 9.80 m/s²O h + t V sin© – g 12/2,where g =9.80 m/s²O h + t V cose - g t2/2,where g = 9.80 m/s²Oh + t V sine – g t212,where g = -9.80 m/s2QUESTION 401 = 26°,02 = 37°,%3DT1 = 89.2 N,g=9.80 m/s2,m = ? kg,Enter your answer using 3 significant figures.T,T,mw

Name: physics - please make it visible QUESTION 3An arrow is shot from a hei
Uploaded: 2024-03-20T06:57:47.000Z
Channel: Bartleby
Description: physics - please make it visible QUESTION 3An arrow is shot from a height h toward a clif.It is shot with a velocity Vat an angle of 8 above the horizontal.It lands on the top edge of the cliff time t later.What is the height of the cliff?O h +t V si

As per our guidelines, we are supposed to answer only first one question in case of multiple…

Science

Physics

QUESTION 3 An arrow is shot from a height h toward a cliff. It is shot with a velocity V at an angle of 0 above the horizontal. It lands on the top edge of the cliff time t later. What is the height of the cliff? O h + t V sin© + 8 1212, where g = 9.80 m/s² O h +t V sin© - g 1212, where g = 9.80 m/s² O h + t V cose – 8 1212, where g = 9.80 m/s2 Oh + t V sine – g 1²12, where g = -9.880 m/s2

QUESTION 3 An arrow is shot from a height h toward a cliff. It is shot with a velocity V at an angle of 0 above the horizontal. It lands on the top edge of the cliff time t later. What is the height of the cliff? O h + t V sin© + 8 1212, where g = 9.80 m/s² O h +t V sin© - g 1212, where g = 9.80 m/s² O h + t V cose – 8 1212, where g = 9.80 m/s2 Oh + t V sine – g 1²12, where g = -9.880 m/s2

Classical Dynamics of Particles and Systems

5th Edition

ISBN:9780534408961

Author:Stephen T. Thornton, Jerry B. Marion

Publisher:Stephen T. Thornton, Jerry B. Marion

Chapter2: Newtonian Mechanics-single Particle

Section: Chapter Questions

Problem 2.6P: In the blizzard of ’88, a rancher was forced to drop hay bales from an airplane to feed her cattle....

See similar textbooks

Similar questions

A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 146 feet per second from an initial height of 89 feet off the ground, then the height of the projectile, hh, in feet, tt seconds after it's shot is given by the equation: h=−16t2+146t+89h=-16t2+146t+89 Find the two points in time when the object is 126 feet above the ground (to one decimal places). Answer: The object is 126 feet off the ground at the following times: (Enter your two answers separated by a comma. Round to ONE decimal places.)
A particle’s velocity along the x-axis is described byv(t) = A t + B t2,where t is in seconds, v is in meters per second, A = 1.06 m/s2, and B = -0.51 m/s3. What is the distance traveled, in meters, by the particle between times t0 = 1.0 s and t1 = 3.0 s?
A ball dropped from rest is traveling 13.0m/s after 0.5s. (a) How far would this ball fall in 15s and (b) where is the ball, the moon (g = 1.6m/s/s), earth (g = 9.8m/s/s), or Jupiter(g = 26.0m/s/s)?
If you take a man to the top of a 20 m tall building and throw him off with a velocity of 5 m/s directed horizontally, what’s the magnitude of his velocity upon hitting the ground (in m/s)? answers include: 22.4 m/s 19.8 m/s 9.8 m/s 20.4 m/s
A cannonball is fired from the ground with anupward initial velocity of 30 m/s. If the cannonball reaches a height of 5 meters, what is its y-velocity? Use g = 10 m/s^2 a. sqrt of 10 m/s b. 10 m/s c. 12 m/s d. 15 m/s
An object is moving in the negative y-direction at 20.0 m/s when it undergoes an acceleration a → = ( 1.00 m / s 2 ) i ^. What is the speed (magnitude of velocity) of the object 10.0 s later?
In the blizzard of ’88, a rancher was forced to drop hay bales from an airplane to feed her cattle. The plane flew horizontally at 160 km/hr and dropped the bales from a height of 80 m above the flat range, (a) She wanted the bales of hay to land 30 m behind the cattle so as to not hit them. Where should she push the bales out of the airplane? (b) To not hit the cattle, what is the largest time error she could make while pushing the bales out of the airplane? Ignore air resistance.
Check your Understanding A cyclist rides 3 km west and then tune around and tides 2 kin east. (a) What Is his displacement? (b) What is the distance traveled? (c) What is the magnitude of his displacement?
A seasoned mini golfer is trying to make par on a tricky number five hole. The golfer can complete the hole by hitting the ball from the flat section it lays on, up a 45∘45∘ ramp, over a moat, and into the hole which is ?=1.90 md=1.90 m away from the end of the ramp. If the opening of the hole and the end of the ramp are at the same height, ?=0.880 m,y=0.880 m, at what speed must the golfer hit the ball to launch the ball over the moat so that it lands directly in the hole? Assume a frictionless surface, so the ball slides without rotating. The acceleration due to gravity is 9.81 m/s2.
A rabbit runs across a parking lot on which a set of coordinate axis has been drawn. The coordinates of the rabbit’s position as a function of time t are given by: x = -0.3t2 + 5t + 28 and y = 0.22t2 - 8t + 30 with t in seconds and x and y in meters. What is the velocity at t= 10 sec? A. 0.6 i + 0.4 j B. -0.2i +1.6j +30 C. -i - 3.6 j D. 4i + 2j
I just wanted to make sure I did this right a) What is the car’s average speed, in km/h, from Hither to Yon for the first part of the trip? G: △d=450Km △T=4.50hr U: v E v=△d△t S v=450km4.50hr S the average speed from Hither to Yon is 100km/h b)How long, in hours, did it take for the car to return to Hither from Yon? G: △d=90.0km/h V=100km/h U: △t E △t=△dV S △t=90.0km100km/h S It took 9.0hr to get back to Hither from Yon c) What is the average velocity of the car for the entire (whole) journey, from Hither to Yon, and back to Hither? G △d =450km + 450km = 900km △T=4.50hr+4.50hr=9.0hr U: V E v=△d△t S v=900km9.0hr S the average velocity for the entire journey is 100km/hr
A ball is thrown with an initial velocity of (30 m/s) at an angle of (60°) above the horizontal, as shown in figure below. Find : (a) The maximum height of the ball? Answer : ( Ymax = 34.5m) (b) The time to rise to the top of the trajectory? Answer:(t,=2.65s) (@) The total time the ball is in the air? Answer: (t:=5.305)