Kinematics Problem: Time to Reach Maximum Height with Initial Velocity 30m/s

[tommy2st]

Homework Statement

A ball is thrown vertically upward from the ground with an initial velocity of 30 m/s. Assume the acceleration due to gravity is g = 10 m s-2.

How long will it take for the ball to rise to the highest point on its trajectory? (AF:no decimals)


i got v1 = 30m/s
a= 10m/s2
x1=0

but i can't think of where to start given this information can someone point me in the right direction

Homework Equations

The Attempt at a Solution

 

[SteamKing]

What will be the velocity of the ball when it reaches the highest point of its trajectory?

 

[tommy2st]

0 is the final velocity. is the answer 3s? i think i may have got it

30=10t
t=3

anyone am i right?

also the height it reaches using 3?

would i use d=30*3 + 1/2(10)(3^2)

 

[tommy2st]

anyone?

 

[Doc Al]

0 is the final velocity. is the answer 3s? i think i may have got it

30=10t
t=3

anyone am i right?

Right!

also the height it reaches using 3?

would i use d=30*3 + 1/2(10)(3^2)

Almost. Careful with signs.

 

[tommy2st]

would it be d= 30*3 + 1/2(-10)(3^2)?

giving me 45?

 

[Doc Al]

would it be d= 30*3 + 1/2(-10)(3^2)?

giving me 45?

Yep!

You can also solve it using the concept of average speed. What's the average speed of the ball as it rises?

 

[tommy2st]

average speed 15 m/s?

also i need help with this part... On the way up, at what height is the speed of the ball equal to half its speed of projection?

i get that it wants me to find at what distance v = 15 but I am having trouble munipulating the equations to solve this

 

[Doc Al]

average speed 15 m/s?

Right. So you can find the distance using average speed X time.

also i need help with this part... On the way up, at what height is the speed of the ball equal to half its speed of projection?

i get that it wants me to find at what distance v = 15 but I am having trouble munipulating the equations to solve this

Solve it exactly as you solved the first part. First figure out the time.

 

[tommy2st]

okay thank you ill try and edit this

t=1.5

d=(15)(1.5)+(1/2)(-10)(1.5^2)= 11.25?

or (30)(1.5)+(1/2)(-10)(1.5^2)= 33.75?

i think its the second one

 

[Doc Al]

okay thank you ill try and edit this

t=1.5

d=(15)(1.5)+(1/2)(-10)(1.5^2)= 11.25?

or (30)(1.5)+(1/2)(-10)(1.5^2)= 33.75?

i think its the second one

Right. The initial velocity is 30 m/s, not 15. (That's the final velocity.)

Verify your answer using the average velocity method.

 

[tommy2st]

thank you so much for your help doc!