The initial velocity is in the y-direction.
A) It's height above ground is
y = x0 + v0*t - 1/2*g*t^2
where y is its height, x0 is its initial position, v0 is it's initial speed in the y direction, g is the acceleration due to gravity and t is time. Plugging in the values from the problem:
y = 6 + 170*t -1/2*g*t^2
b) The maximum height will be found at the point when dy/dt equals zero
dy/dt = 170 - 1/2*g*t^2 = 0
Solve for t, then plug this value back into y to find the maximum height
c) The rock will hit the ground when y = 0
0 = 6 + 170*t -1/2*g*t^2
Solve for t using the quadratic equation solutions
d) the velocity is dy/dt = 170 - 1/2*g*t^2
Plug your answer for t from part c into this equation to find the velocity when it hits the ground
aj uses a lingshot to launch a rock straight up from the a point of 6 ft. above ground level with an initial velocity of 170 ft per second.
A) find an equation that models the height of the rock t seconds after it is launched.
b) what is the maximum height of the rock and when will it reach it. explain how you found this answer or show work.
c)when will the rock hit the ground? explain how you found this answer or show work.
d) what is the rock's vertical velocity when it hits the ground? show work
1 answer
3 answers