Ball thrown upward given time

[sbright]

Homework Statement

Find initial speed and maximum height of a ball that is thrown 90 degrees straight up and lands on the ground in 3 seconds. What happens when the ball is thrown at some angle other than 90 degrees? Does it change the time of flight? Can you still determine the max height and initial speed?

Homework Equations

v=vo+at
Vy=Vsin90
y=vot+1/2gt^2

The Attempt at a Solution

For initial speed i did 9.81*3=29.43 to find velocity then (29.43)sin90=29.43 which i believe is the initial speed(Vo)
I'm having trouble with max height.
(29.43)(3)+4.9(3^2)=132.39m
Which i believe is much to big of a number..help please
Also I think as you move closer to 0 degrees and further away from 90 degrees, time of flight decreases, but can we determine max height and initial speed?

 

[PeterO]

Homework Statement

Find initial speed and maximum height of a ball that is thrown 90 degrees straight up and lands on the ground in 3 seconds. What happens when the ball is thrown at some angle other than 90 degrees? Does it change the time of flight? Can you still determine the max height and initial speed?

Homework Equations

v=vo+at
Vy=Vsin90
y=vot+1/2gt^2

The Attempt at a Solution

For initial speed i did 9.81*3=29.43 to find velocity then (29.43)sin90=29.43 which i believe is the initial speed(Vo)
I'm having trouble with max height.
(29.43)(3)+4.9(3^2)=132.39m
Which i believe is much to big of a number..help please
Also I think as you move closer to 0 degrees and further away from 90 degrees, time of flight decreases, but can we determine max height and initial speed?

Firstly, this ball only traveled up for 1.5 seconds, then down for 1.5 seconds which is why it landed after 3 seconds.

 

[sbright]

Firstly, this ball only traveled up for 1.5 seconds, then down for 1.5 seconds which is why it landed after 3 seconds.

So would this be correct
29.43(1.5)+4.9(1.5^2)=55.17
still seems like too big of a number...

 

[sbright]

Ok so I'm pretty sure I found the right answer
29.43(1.5)-4.9(1.5^2)=33.12m

Now if anyone could help with the very last question, "What happens when you throw a ball at some angle other than 90 degrees? Can you still determine the maximum height and initial speed? Give careful explanations in your answers.

Any input with that last question would make me the happiest person in the world thank you!

 

[sbright]

Remember* only given time (for that last question)

 

[PeterO]

Ok so I'm pretty sure I found the right answer
29.43(1.5)-4.9(1.5^2)=33.12m

Now if anyone could help with the very last question, "What happens when you throw a ball at some angle other than 90 degrees? Can you still determine the maximum height and initial speed? Give careful explanations in your answers.

Any input with that last question would make me the happiest person in the world thank you!

I find the follow up a little ambiguous.

What happens when the ball is thrown at some angle other than 90 degrees? Does it change the time of flight? Can you still determine the max height and initial speed?


You are asked "does it change the time of flight?"
Well if the ball still lands 3 seconds later, clearly no. However if we are to assume that it was thrown at the same speed you calculate in the first part, then clearly yes.

The maximum height is always determined by the flight time.
If an object is in the air for 3 seconds, it travels up for 1.5 seconds and down for 1.5 seconds. How far down does anything fall in 1.5 seconds [always ignoring iar resistance]

 

[PeterO]

Ok so I'm pretty sure I found the right answer
29.43(1.5)-4.9(1.5^2)=33.12m

Now if anyone could help with the very last question, "What happens when you throw a ball at some angle other than 90 degrees? Can you still determine the maximum height and initial speed? Give careful explanations in your answers.

Any input with that last question would make me the happiest person in the world thank you!

That answer is far too high, since your initial velocity was based on 3 seconds to each maximum height, and it is only 1.5 seconds.