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anonymous
 5 years ago
A projectile is fired straight up. y"=GM/((Y+R)^2) Be^y y(0)=0 y'(0)=V_0 Compute total amount of time for the projectile to stay in the sky from launch till retouch of the ground. G, M, R, V_0, and B are all constants.
Compute the total amount of time the projectile is in the sky from launch to retouch of the ground, and compute the velocity at the instant when the projectile retouches the ground.
Am i supposed to integrate y" twice to find the position and set y=0 in the position function? how do i integrate the acceleration (which is a function of time) with respect to time?
anonymous
 5 years ago
A projectile is fired straight up. y"=GM/((Y+R)^2) Be^y y(0)=0 y'(0)=V_0 Compute total amount of time for the projectile to stay in the sky from launch till retouch of the ground. G, M, R, V_0, and B are all constants. Compute the total amount of time the projectile is in the sky from launch to retouch of the ground, and compute the velocity at the instant when the projectile retouches the ground. Am i supposed to integrate y" twice to find the position and set y=0 in the position function? how do i integrate the acceleration (which is a function of time) with respect to time?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you're headed in the right direction.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have two given initial values for y(0) and y'(0), so you can solve for each arb constant when integrating.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do i integrate acceleration (which is a function of time) with respect to time?how would you integrate f''(x)=x^2+x^3 with respect to x? Same way right? f is a function of x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so, you'll eventually get a position function: y(t)...should be able to solve for 0 & get the time interval that the projectile is in flight...velocity at the instant the projectile hits the ground is zero...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but just before it hits the ground, it (the velocity) will have an x & y component (a magnitude and direction)...you probably know how to find those ; )
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