Finding the initial velocity of a thrown object using its time to max height

[MaronKun]

Homework Statement

An object is thrown from the ground into the air at an angle of 30 degrees to the horizontal. If this object reaches a maximum height of 5.75m, at What velocity was it thrown?

a= -9.81 m/s
h= 5.75 m
I know that velocity is the one that I need to solve but I'm kinda finding it hard for myself to cope up how to solve for the time because I know that once I figure out the time, I would easily be able to solve for the velocity

Homework Equations

• a= Vf-vi/t
• t=0-vi/a
• t= -vi/a

The Attempt at a Solution

• a= Vf-vi/t
• t=0-vi/a
• t= -vi/a
• T= -sin(30/-9.81
• T=0.05s

I know I probably used the wrong solution but I'm running out of option. I tried googling it but I can't find an answer that I could actually grasp in my mind. I have a quiz tomorrow and I'm afraid there might be a question like this and I might have a hard time figuring it out if I don't learn it by tonight. I hope you guys can help me. Sorry for the messed up format because I'm not really used to the settings here in forums :).

 

[Simon Bridge]

I know that velocity is the one that I need to solve but I'm kinda finding it hard for myself to cope up how to solve for the time because I know that once I figure out the time, I would easily be able to solve for the velocity

... hint: conservation of energy.
BTW: what is sin(30deg)?

 

[haruspex]

• t= -v_i/a

• T= -sin(30)/-9.81

What happened to v_i?
There are 5 standard variables in the 5 SUVAT equations, each involving four of the variables.
In any constant acceleration problem, identify the three variables whose value you know and a fourth whose value is to be determined. Pick the equation that involves those four.
What other SUVAT equations do you know? (As Simon says, one of them corresponds to conservation of mechanical energy.)

 

[MaronKun]

... hint: conservation of energy.
BTW: what is sin(30deg)?

I'm not really sure. I had a similar question like this on my textbook but instead of height is the given, it was the x displacement. My teacher wrote a solution of T=D/V > 8.59/v(cos)23
then he substituted that for the time in D=Vit + .5at^2 which became D=vsin23(8.59/cos23) +0.5(9.81)(8.59/cos23)^2 which eventually gave the velocity for the horizontal. I just assumed that the 8.59/v(cos)23 to solve for time would have been similar to this question.

 

[Simon Bridge]

Trying to copy what was done in class without understanding it may work but usually leads to error.
Do you know what energy transformations are involved?
Do you know how to separate the motion into vertical and horizontal components?

 

[cotton_field]

The answer should be 21.2m/s

v^2 = u+2ad
v^2=2(-9.81)(5.75)
√(v^2=112.815)
v=10.62144058
Then use the a=(v-u)/t formula, we would only need the (v-u)part
-10.62-10.62 = -21.24288116
The velocity must be positive in this case and due to the sig dig therefore the answer is 21.2m/s you could use the angle of 30 degrees to check your answer.
sin(30)=y/21.2 cos(30)=x/21.2
y= 10.6 x= 18.39687473

 

[berkeman]

The answer should be 21.2m/s

v^2 = u+2ad
v^2=2(-9.81)(5.75)
√(v^2=112.815)
v=10.62144058
Then use the a=(v-u)/t formula, we would only need the (v-u)part
-10.62-10.62 = -21.24288116
The velocity must be positive in this case and due to the sig dig therefore the answer is 21.2m/s you could use the angle of 30 degrees to check your answer.
sin(30)=y/21.2 cos(30)=x/21.2
y= 10.6 x= 18.39687473

Welcome to PF.

Please keep in mind that we do not allow providing solutions to homework in the Homework Help forums (that's in the PF Rules -- see INFO at the top of the page). However, since this thread is 8 years old, it is generally okay to post a solution this late after the student has moved on.

I haven't checked your solution, BTW.

 

[PeroK]

I haven't checked your solution, BTW.

Well, ##u_y^2 = 2gh## and ##u_y = u\sin \theta##. So:$$u = \frac{\sqrt{2gh}}{\sin \theta}$$And, with ##\sin(30)=\frac 1 2##:
$$u = 2\sqrt{2gh} = 21.2 \ m/s$$