SOLUTION: When a ball is thrown upward at a speed of 19 m/s, its height s above the ground (in meters) after t seconds is given by the formula s=19t-4.9t^2 Find the height of the ball af

Question 483292: When a ball is thrown upward at a speed of 19 m/s, its height s above the ground (in meters) after t seconds is given by the formula
s=19t-4.9t^2
Find the height of the ball after 3 seconds
Found 2 solutions by bucky, mathstutor494:
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
You are given that the height of a ball that is thrown upward with an initial velocity of 19m/s can be calculated by using the equation:
.

.
where t is the number of seconds after the ball is released.
.
You are now asked to find the height of the ball after 3 seconds. You can do this simply by substituting 3 for t in the equation as follows:
.

.
First multiply the 19 times 3 and you get 57. Substitute this into the equation and you have:
.

.
Next square the 3 by recognizing that is and that is 9. So substitute 9 for and the equation then becomes:
.

.
Multiply 4.9 times 9 and get 44.1
.
Substitute this into the equation and you have:
.

.
Finally, subtract the 44.1 from 57 and you have the answer:
.

.
Don't forget that the answer is in the units of meters. So after throwing the ball with an initial velocity of 19 m/s, after 3 seconds the ball is 12.9 meters above the height at which is was released.
.
As a little bit of an explanation of what this equation means, recognize that distance is rate (the 19 m/s) times time. So if gravity didn't work, you launch the ball upward and the distance it would be after t seconds is given by 19 times t.
.
However, gravity does work on the ball. The effect of gravity is given by the second term in the equation. Since gravity is working downward, it has a negative or downward effect on the ball and that's why the second term in the equation has a negative or minus sign.
.
If gravity didn't work, 3 seconds after launch, the ball would have been 57 meters above the launch point. But the effect of gravity has taken away 44.1 meters of the height ... that's the . And the result is that the ball is only 12.9 meters above the launch height.
.
The math simply provides a way to help you understand and analyze the physics of the situation.
.
Good luck and I hope this helps.
.
Answer by mathstutor494(120) (Show Source): You can put this solution on YOUR website!
Height of the ball after 3 seconds is achieved by substituting t=3 in below formula
s=19t-4.9t^2
s=19*3-4.9*3^2
s= 57 - 4.9*9
s= 57- 44.1
s= 12.9 m

RELATED QUESTIONS
When a ball is thrown upward at a speed of 19m/s, its height s above the ground (in... (answered by Alan3354)
A ball thrown in the air with a velocity of 24.6 m/s is at a height represented by... (answered by stanbon)
help me please A ball is thrown from a height of 45 meters with an initial downward... (answered by ewatrrr)
A ball is thrown upward. The path of the ball is modelled by the equation: h=-4.9t^2 +... (answered by Alan3354)
a ball is thrown vertically upward with an initial speed of 48 ft/s. its height, in feet, (answered by richwmiller)
a ball is thrown vertically upward with an initial speed of 48 ft/s. its height, in feet, (answered by ewatrrr)
Applications of Quadratic Functions 1. The hypotenuse of a right triangle is 32 feet... (answered by ikleyn)
A ball thrown vertically into the air has its height above ground given by h=-16t^2+112t, (answered by nerdybill)
A ball is thrown vertically upward with an intial velocity of 96 feet per second. The... (answered by Alan3354)