.
To calculate the height of the object 7 seconds after it leaves the ground,
simply substitute t= 7 into the given formula for h(t) and calculate.
You will get then
h(7) = -16*7^2 + 32*7 + 5 = -555.
It means that at t= 7 seconds the ball will be much lower than the ground surface,
which means that at t= 7 the formula DOES NOT work, actually.
At the time t=7 seconds the ball will be, gm-m-m, on the ground surface or somewhere else,
but the given formula just does not work for this time moment.
To answer the second question, you should solve this equation
h(t) = 0 = -16*t^2 + 32t + 5 = 0.
I leave it to you to solve it - it is a simple exercise.
The last notice is that the formula for h(t) works only till that moment,
when the ball will reach the ground.
Is everything clear to you from my explanation ?
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In this site, there is a bunch of lessons on a projectile thrown/shot/launched vertically up
- Problem on a projectile moving vertically up and down
- Problem on an arrow shot vertically upward
- Problem on a ball thrown vertically up from the top of a tower
- Problem on a toy rocket launched vertically up from a tall platform
Consider these lessons as your textbook, handbook, tutorials and (free of charge) home teacher.
Read them attentively and learn how to solve this type of problems once and for all.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this textbook under the topic "Projectiles launched/thrown and moving vertically up and dawn".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.
Happy learning (!)