Homework Help based on the equations of motion

  • #1

Homework Statement


A particle is projected vertically upwards at 30 m/s. Calculate (a) how long it takes to reach its maximum height, (b) the two times at which it is 40 m above the point of projection, (c) the two times at which it is moving at 15 m/s.


2. Homework Equations [/b
final velocity (v)
initial velocity (u)
acceleration due to gravity (a) = 10m/s^2
time (t)
Distance or displacement (s)
v^2=u^2 + 2as
v=u+at
s=ut+1/2 at^2
s=1/2(u+v)t


The Attempt at a Solution


I first find the maximum height, this is where i used the equation v^2=u^2 +2as
0^2=30^2 + 2(10)s
s=45 m maximum height

then i used that value and plugged into the equation to find time
45= 1/2 (30+0)t
t= 3seconds

therefore total time is 3 seconds for the particle to reach its maximum height.

part (b) is where i am stuck

however i move on to part (c)
and i plugged in the values in the equation
15=30-10t
t=1.5 seconds

however the question asked for two (2) times. so i am also stuck with finding the second time...
 

Answers and Replies

  • #2
berkeman
Mentor
59,054
9,156

Homework Statement


A particle is projected vertically upwards at 30 m/s. Calculate (a) how long it takes to reach its maximum height, (b) the two times at which it is 40 m above the point of projection, (c) the two times at which it is moving at 15 m/s.


2. Homework Equations [/b
final velocity (v)
initial velocity (u)
acceleration due to gravity (a) = 10m/s^2
time (t)
Distance or displacement (s)
v^2=u^2 + 2as
v=u+at
s=ut+1/2 at^2
s=1/2(u+v)t


The Attempt at a Solution


I first find the maximum height, this is where i used the equation v^2=u^2 +2as
0^2=30^2 + 2(10)s
s=45 m maximum height

then i used that value and plugged into the equation to find time
45= 1/2 (30+0)t
t= 3seconds

therefore total time is 3 seconds for the particle to reach its maximum height.

part (b) is where i am stuck

however i move on to part (c)
and i plugged in the values in the equation
15=30-10t
t=1.5 seconds

however the question asked for two (2) times. so i am also stuck with finding the second time...


Welcome to the PF.

When they are asking for 2 times, one will be on the way up, and the other on the way down. There is a squared term in the equation for the vertical motion, and you should be able to use both roots in solving for the two times. Does that help?
 
  • #3
no sorry im lost....
 
  • #4
gneill
Mentor
20,913
2,862
no sorry im lost....
When the problem says, "...it is moving at 15 m/s", this is giving you a value for the speed of the projectile.

A speed of 15 m/s can mean the projectile is going up (+15 m/s) or down (-15 m/s). Speeds don't have directions associated with them, velocities do :wink:
 

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