PLEASE USE FORMULAS AND BOX YOUR ANSWERS.Thanks! Section 4 You are outside in a

Question

PLEASE USE FORMULAS AND BOX YOUR ANSWERS.Thanks!

Section 4

You are outside in a big, level field, in the midst of which stands a narrow wall, 14.8m high. You are given a device that can launch a projectile, always with the same speed of 42.0m/s

A) first, you launch the projectile form the top of the wall, at an angle of 51.0 degrees above the horizontal. How far from the base of the wall does the projectile land on the ground? __________.______m

next, you launch another projectile from the top of the wall. After 3.1seconds of flight, the projectile lands on the ground.

B) At what angle above the horizontal was the projectile launched this time? _________.______ degrees

C) How far from the base of the wall does the projectile land in this case? ________._____m

Now you position the launcher on the ground, a distance of 20.3m from the base of the wall. You launch the projectile directly toward the wall, at an angle of 16.8 degrees above the horizontal.

D) How high above the ground does the projectile strike the wall? ________._____ m

E) stil positioned at the same distance 20.3m from the base of the wall, what is the min launch angle with respect to the horizontal that will allow the projectile to just make it over the top of the wall? _________.____ degrees

F) what is the maximum disance from the bottom of the wall that you can position your launcher on the ground, such that it's possible to make it over the wall? ______.____m

Explanation / Answer

vertically upward direction will be considered positive.

hence acceleration due to gravity=-9.8 m/s^2

part A:

initial vertical speed=42*sin(51)=32.64 m/s

vertical displacement=-14.8 m

if time taken is t seconds

then -14.8=32.64*t-0.5*9.8*t^2

==>4.9*t^2-32.64*t-14.8=0

solving for t , we get t=7.0874 seconds

then distance from the base of the wall=horizontal speed*time

=42*cos(51)*7.0874=187.33 m

part B:

let initial vertical speed be v.

then -14.8=v*3.1-0.5*9.8*3.1^2

==>v=(4.9*3.1^2-14.8)/3.1=10.416 m/s

hence if angle above horizontal is theta,

then 42*sin(theta)=10.416

==>theta=14.359 degrees

part C:

distance from base=42*cos(10.416)*3.1=128 m

part D:

distance to be covered=20.3 m

horizontal speed=42*cos(16.8)=40.207 m/s

time taken=20.3/40.207=0.50488 seconds

then height above ground=42*sin(16.8)*0.50488-0.5*9.8*0.50488^2=4.88 m

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