Homework Help based on the equations of motion

[risingabove]

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...

 

[berkeman]

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?

 

[risingabove]

no sorry I am lost...

 

[gneill]

no sorry I am 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

 

[Nickie]

Your approach to finding the maximum height and time is correct. For part (b), you can use the equation s=ut+1/2at^2 to find the two times at which the particle is 40m above the point of projection. Since the initial velocity is 30 m/s and acceleration is -10 m/s^2 (due to gravity), the equation becomes:

40=30t+1/2(-10)t^2

Solving for t gives two values, t=1 second and t=4 seconds. Therefore, the particle is 40m above the point of projection at 1 second and 4 seconds.

Similarly, for part (c), you can use the equation v=u+at to find the two times at which the particle is moving at 15 m/s. Plugging in the values, the equation becomes:

15=30-10t

Solving for t gives two values, t=1.5 seconds and t=3 seconds. Therefore, the particle is moving at 15 m/s at 1.5 seconds and 3 seconds.

Remember to always check your units and make sure they are consistent throughout the calculations. Also, it is helpful to draw a diagram or visualize the motion to better understand the problem. Great job on your attempt and keep up the good work!