NASA launches a rocket at t = is given by h(t) = - 4.912 + 349t + 318. O seconds. Its height, in meters above sea-level, as a function of time Assuming that the rocket will splash down into the ocean, at what time does splashdown occur? The rocket splashes down after seconds. How high above sea-level does the rocket get at its peak? The rocket peaks at meters above sea-level.

NASA launches a rocket at t = is given by h(t) = - 4.912 + 349t + 318. O seconds. Its height, in meters above sea-level, as a function of time Assuming that the rocket will splash down into the ocean, at what time does splashdown occur? The rocket splashes down after seconds. How high above sea-level does the rocket get at its peak? The rocket peaks at meters above sea-level.

Name: NASA launches a rocket at t =is given by h(t) = - 4.912 + 349t + 318.
Uploaded: 2024-03-05T11:56:34.000Z
Channel: Bartleby
Description: NASA launches a rocket at t =is given by h(t) = - 4.912 + 349t + 318.O seconds. Its height, in meters above sea-level, as a function of timeAssuming that the rocket will splash down into the ocean, at what time does splashdown occur?The rocket splas

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter3: Motion In Two Dimensions

Section: Chapter Questions

Problem 6P: At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of...

See similar textbooks

Similar questions

ASA launches a rocket at t=0t=0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t)=−4.9t2+226t+227hAssuming that the rocket will splash down into the ocean, at what time does splashdown occur?The rocket splashes down after seconds.How high above sea-level does the rocket get at its peak?The rocket peaks at meters above sea-level.
A fly is buzzing through the air at 0.1 m/d when a gust of wind comes and gives it an acceleration of 0.2 m/s^2 for 4.7 seconds. How far has the fly traveled during this time frame, being blown forward?
A 238U nucleus is moving in the x direction at 5.0×105 m/s when it decays into an alpha particle (4He) and a 234Th nucleus. If the alpha particle moves off at 22 degrees above the x axis with a speed of 1.1×107 m/s, a) What is the speed of the thorium nucleus and b) What is the direction of the motion of the thorium nucleus ( degrees clockwise from the x axis)?
An object thrown vertically upward from the surface of a celestial body at a velocity of24 m/s reaches a height of s=−0.3t^2+24t meters in t seconds. d. When does the object strike the ground? e. With what velocity does the object strike the ground? f. On what intervals is the speed increasing?
Two runners start simultaneously from the same point on a circular 200-m track and run in opposite directions. One runs at a constant speed of 6.20 m/s, and the other runs at a constant speed of 5.50 m/s. When they first meet, (a) for how long a time will they have been running, and (b) how far will each one have run along the track?
How far from the origin has the particle gone after 0.11 minutes if Vx = 3m/s , Vy = 2 m/s and Vz = 3.31 m/s. Assume that the particle is moving at a constant rate.
A bullet is traced with a fixed ruler and flash twice witha time interval of 2 ± 0.06 ms. If the position of the bullet determined is 23.0 ± 0.5 cm firstand the position of the bullet is 37.5 ± 0.5 cm next. Calculate the speed of the bullet, V(=D/T).
A jeep travels a distance d=22.1m in the positive x direction in a time t1=20.2s, at which point the jeep brakes, coming to rest in t2=7.38s. 1. What was the jeep's instantaneous velocity in the horizontal direction, in meters per second, when it began braking? 2. Using the result from question 1, what was the jeep's horizontal component of acceleration, in meters per squared second, during the braking period?
If an object is located 20m to the right of the origin at 1:00 PM, and later the object is located 30 m to the left of the origin at 2:00 PM, then the displacement from 1:00 PM to 2:00 PM is what?
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, h, in feet, t 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 moving along a line has position s(t) = t4 − 18t2 at time t seconds. At which times does the particle pass through the origin? At which times is the particle instantaneously motionless (i.e., it has zero velocity)?
An athlete crosses a 29 m wide river by swimming perpendicular to the water current at a speed of 1.1 m/s relative to the water. He reaches the opposite side at a distance 35m downstream from his starting point. How fast is the water in the river flowing with respect to the ground in m/s?