(a) Find the time when the stone reaches its maximum height.
Write the velocity and position kinematic equations.
v = at + v0
Δy = y - y0 = v0t + at^2
Substitute a = -9.80 m/s^2, v0 = 21.9 m/s, and y0 = 0 into the preceding two equations.
v = (-9.80 m/s^2)t + 21.9 m/s (1)
y = (21.9 m/s)t - (4.90 m/s^2)t^2 (2)
Substitute v = 0, the velocity at maximum height, into Equation (1) and solve for time.
0 = (-9.80 m/s^2)t + 20.0 m/s
(b) Determine the stone's maximum height.
Substitute the time found in part (a) into Equation (2).
ymax = (21.9 m/s)(t s) - (4.90 m/s^2)(t s)^2 = m
(c) Find the time the stone takes to return to its initial position, and find the velocity of the stone at that time.
Set y = 0 in Equation (2) and solve for t.
0 = (21.9 m/s)t - (4.90 m/s^2)t^2
= t(21.9 m/s - 4.90 m/s^2t)
t = s
Substitute the time into Equation (1) to get the velocity.
v = 21.9 m/s + (-9.80 m/s^2)(t)
v = m/s
(d) Find the time required for the stone to reach the ground.
In Equation (2), set y = -57.7 m.
-57.7 m = (-21.9 m/s)t - (4.90 m/s^2)t^2
Apply the quadratic formula and take the positive root.
t = s
(e) Find the velocity and position of the stone at t = 6.68 s. Substitute values into Equations (1) and (2).
v = (-9.80 m/s^2)(6.68 s) + 21.9 m/s
v = m/s
y = (21.9 m/s)(6.68 s) - (4.90 m/s^2)(6.68 s)^2
y = m
A Projectile is launched straight up at 56.5 m/s from a height of 88.5 m, at the edge of a sheer cliff. The projectile falls just missing the cliff and hitting the ground below.
(a) Find the maximum height of the projectile above the point of firing.
(b) Find the time it takes to hit the ground at the base of the cliff.
(c) Find its velocity at impact. m/s