4. A projectile has an initial velocity of 20.0 m/s at an angle of 35.0° with respect to the x-direction. The object is initially 50.0 m above ground level. a) What is the maximum height of the object relative to the ground level? Vi= 20m is 30: 16.4mis 35° Vxi = 20 Cos (35) Vyi = 20 sin (35°) Vy=0 @ AYMAX Vyf? = Vyi"-2gay =11. Smis %3D %3D 50m 2. %3D hmax = 0=(11.5)²-29AY %3D = Aymax +h -13 2.25= - 29Ay %3D 6.75m + Som -132.25= AY 56.8 m 19.6 Ay = 6.75 m b) What total time is the object in the air for? Źğt²+ Vyit 2. Ay %3D - 56.8 = -gt 2. |1. 5t 11. 5t/s 9.9t+11.5t + 56.8=0 %3D A = -4.9 B = ||.5 C = 56.8 %3D %3D t= -|1.5t V(11.5)²_4(-4.9)(56.8) 21-4.9) t = - 11.5+ 13 2.25 + 1|1328 -11.5t -9.8 t = - 11. S t 35.292 4.773 t=7.478

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter3: Motion In Two Dimensions
Section: Chapter Questions
Problem 39AP: A rocket is launched at an angle of 53.0 above the horizontal with an initial speed of 100. m/s. The...
icon
Related questions
Question

Answer part B). What total time is the object in the air for?

4. A projectile has an initial velocity of 20.0 m/s at an angle of 35.0° with respect
to the x-direction. The object is initially 50.0 m above ground level.
a) What is the maximum height of the object relative to the ground level?
Vi= 20m is
30:
16.4mis
35°
Vxi = 20 Cos (35)
Vyi = 20 sin (35°)
Vy=0
@ AYMAX
Vyf? = Vyi"-2gay
=11. Smis
%3D
%3D
50m
2.
%3D
hmax =
0=(11.5)²-29AY
%3D
= Aymax +h
-13 2.25= - 29Ay
%3D
6.75m + Som
-132.25= AY
56.8 m
19.6
Ay = 6.75 m
b) What total time is the object in the air for?
Źğt²+ Vyit
2.
Ay
%3D
- 56.8 = -gt
2.
|1. 5t
11. 5t/s
9.9t+11.5t + 56.8=0
%3D
A = -4.9
B = ||.5
C = 56.8
%3D
%3D
t= -|1.5t V(11.5)²_4(-4.9)(56.8)
21-4.9)
t = - 11.5+ 13 2.25 + 1|1328
-11.5t
-9.8
t = - 11. S t 35.292
4.773
t=7.478
Transcribed Image Text:4. A projectile has an initial velocity of 20.0 m/s at an angle of 35.0° with respect to the x-direction. The object is initially 50.0 m above ground level. a) What is the maximum height of the object relative to the ground level? Vi= 20m is 30: 16.4mis 35° Vxi = 20 Cos (35) Vyi = 20 sin (35°) Vy=0 @ AYMAX Vyf? = Vyi"-2gay =11. Smis %3D %3D 50m 2. %3D hmax = 0=(11.5)²-29AY %3D = Aymax +h -13 2.25= - 29Ay %3D 6.75m + Som -132.25= AY 56.8 m 19.6 Ay = 6.75 m b) What total time is the object in the air for? Źğt²+ Vyit 2. Ay %3D - 56.8 = -gt 2. |1. 5t 11. 5t/s 9.9t+11.5t + 56.8=0 %3D A = -4.9 B = ||.5 C = 56.8 %3D %3D t= -|1.5t V(11.5)²_4(-4.9)(56.8) 21-4.9) t = - 11.5+ 13 2.25 + 1|1328 -11.5t -9.8 t = - 11. S t 35.292 4.773 t=7.478
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Knowledge Booster
Distance and Speed
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning