Question #c287b
1 Answer
(A)
(B)
(C)
Explanation:
We're asked to find
-
(A) the maximum height the stone reaches
-
(B) the velocity at which the stone strikes the ground
-
(C) the time taken to reach the maximum height
- (A) Maximum Height
To find the maximum height obtained by the stone (and all other information we need), we'll first find the initial velocity of the stone, using the equation
ul(y = y_0 + v_(0y)t - 1/2g t^2
where
-
y is the final position (0 , ground level) -
y_0 is the initial position (85 "m" ) -
v_(0y) is the initial velocity (what we're trying to find) -
t is the time (5 "s" ) -
g = 9.81 "m/s"^2
Plugging in known values:
0 = 85 "m" + v_(0y)(5color(white)(l)"s") - 1/2(9.81color(white)(l)"m/s"^2)(5color(white)(l)"s")^2
1/2(9.81color(white)(l)"m/s"^2)(5color(white)(l)"s")^2 = 85color(white)(l)"m" + v_(0y)(5color(white)(l)"s")
color(red)(v_(0y)) = (1/2(9.81color(white)(l)"m/s"^2)(5color(white)(l)"s")^2 - 85color(white)(l)"m")/(5color(white)(l)"s") = color(red)(ul(7.53color(white)(l)"m/s"
At the stone's maximum height, its instantaneous
ul((v_y)^2 = (v_(0y))^2 - 2g(y - y_0)
The variables here we know except for the final height
0 = (color(red)(7.53color(white)(l)"m/s"))^2 - 2(9.81color(white)(l)"m/s"^2)(y - 85color(white)(l)"m")
color(blue)(ulbar(|stackrel(" ")(" "y = 87.9color(white)(l)"m"" ")|)
(B) Final Velocity
To find the final velocity, we can use the equation
ul(v_y = v_(0y) - g t
where
-
v_y is the final velocity (what we're trying to find) -
v_(0y) is the initial velocity (found to becolor(red)(7.53color(white)(l)"m/s" ) -
g = 9.81 "m/s"^2 -
t = 5 "s"
Plugging these in:
color(blue)(v_y) = color(red)(7.53color(white)(l)"m/s") - (9.81color(white)(l)"m/s"^2)(5color(white)(l)"s") = color(blue)(ulbar(|stackrel(" ")(" "-41.5color(white)(l)"m/s"" ")|)
- (C) Time at Maximum Height
To find the time when the stone reaches its maximum height, we can again use the equation
ul(v_y = v_(0y) - g t
where
-
v_y = 0 (at maximum height) -
v_(0y) = color(red)(ul(7.53color(white)(l)"m/s" -
g = 9.81 "m/s" -
t is the time (what we're trying to find)
And so we have
0 = color(red)(7.53color(white)(l)"m/s") - (9.81color(white)(l)"m/s"^2)t
color(blue)(ulbar(|stackrel(" ")(" "t = 0.767color(white)(l)"s"" ")|)