High School Calculus Velocity question

Noriko Kamachi
Messages
6
Reaction score
0

Homework Statement


If you throw a ball straight down from a building that is 443 meters tall with a velocity of 22 m/s how long would it take to reach the ground, and what would be its speed at impact?

Homework Equations


Since the ball is being thrown down with a velocity of 22 m/s I plug that into the gravitational constant equation for Earth -4.9t^2 -22t +443.

The Attempt at a Solution


To get when the ball hits the ground I set -4.9t^2 -22t +443=0. Solving for t I get about ~7.5. To get the speed at impact I took the second derivative which is -9.8t -22. Plugging in t into this equation I get |-9.8(7.5) -22| for 95.5 m/s for the speed at impact. Was this problem handled in the correct fashion?
 
Physics news on Phys.org
Noriko Kamachi said:

Homework Statement


If you throw a ball straight down from a building that is 443 meters tall with a velocity of 22 m/s how long would it take to reach the ground, and what would be its speed at impact?

Homework Equations


Since the ball is being thrown down with a velocity of 22 m/s I plug that into the gravitational constant equation for Earth -4.9t^2 -22t +443.

The Attempt at a Solution


To get when the ball hits the ground I set -4.9t^2 -22t +443=0. Solving for t I get about ~7.5. To get the speed at impact I took the second derivative which is -9.8t -22. Plugging in t into this equation I get |-9.8(7.5) -22| for 95.5 m/s for the speed at impact. Was this problem handled in the correct fashion?

It looks good to me.
 
  • Like
Likes Noriko Kamachi
Noriko Kamachi said:

Homework Statement


If you throw a ball straight down from a building that is 443 meters tall with a velocity of 22 m/s how long would it take to reach the ground, and what would be its speed at impact?

Homework Equations


Since the ball is being thrown down with a velocity of 22 m/s I plug that into the gravitational constant equation for Earth -4.9t^2 -22t +443.

The Attempt at a Solution


To get when the ball hits the ground I set -4.9t^2 -22t +443=0. Solving for t I get about ~7.5. To get the speed at impact I took the second derivative which is -9.8t -22. Plugging in t into this equation I get |-9.8(7.5) -22| for 95.5 m/s for the speed at impact. Was this problem handled in the correct fashion?
-9.8t-22 is not the second derivative; it's the first derivative of the position function, -4.9t2-22t+443 = 0.
 
  • Like
Likes Noriko Kamachi
Dick said:
It looks good to me.

Thanks! :D

SteamKing said:
-9.8t-22 is not the second derivative; it's the first derivative of the position function, -4.9t2-22t+443 = 0.

Ah I see! Thanks for correcting me!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top