Question

Problem A stone is thrown from the top of a building with an initial velocity of 24.3 m/s straight upward, at an initial height of 53.1 m above the ground. The stone just misses the edge of the roof on its way down, as shown in Figure 2.20.

(a) Determine the time needed for the stone to reach its maximum height.
(b) Determine the maximum height.
(c) Determine the time needed for the stone to return to the height from which it was thrown, and the velocity of the stone at that instant.
(d) Determine the time needed for the stone to reach the ground.
(e) Determine the velocity and position of the stone at t = 5.68 s.

(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 +  at2
Substitute a = -9.80 m/s2, v0 = 24.3 m/s, and y0 = 0 into the preceding two equations.	v = (-9.80 m/s2)t + 24.3 m/s  (1) y = (24.3 m/s)t - (4.90 m/s2)t2  (2)
Substitute v = 0, the velocity at maximum height, into Equation (1) and solve for time.	0 = (-9.80 m/s2)t + 20.0 m/s s
(b) Determine the stone's maximum height.
Substitute the time found in part (a) into Equation (2).	ymax = (24.3 m/s)(t s) - (4.90 m/s2)(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 = (24.3 m/s)t - (4.90 m/s2)t2 = t(24.3 m/s - 4.90 m/s2t) t =  s
Substitute the time into Equation (1) to get the velocity.	v = 24.3 m/s + (-9.80 m/s2)(t) v =  m/s
(d) Find the time required for the stone to reach the ground.
In Equation (2), set y = -53.1 m.	-53.1 m = (-24.3 m/s)t - (4.90 m/s2)t2
Apply the quadratic formula and take the positive root.	t =  s
(e) Find the velocity and position of the stone at t = 5.68 s. Substitute values into Equations (1) and (2).	v = (-9.80 m/s2)(5.68 s) + 24.3 m/s v =  m/s y = (24.3 m/s)(5.68 s) - (4.90 m/s2)(5.68 s)2 y =  m

Answer 1