A stone from the catapult is thrown upward with an initial velocity v0 = 45 m / s. a) find the maximum lift

A stone from the catapult is thrown upward with an initial velocity v0 = 45 m / s. a) find the maximum lift b) how long will the stone be in flight?

V0 = 45 m / s.

g = 10 m / s2.

hmax -?

t -?

The stone moves under the influence of gravity, so it moves with the acceleration of gravity g. First, it decelerates to a complete stop V = 0 m / s at the maximum height hmax, and then moves downward until it hits the ground.

h = (V0 ^ 2 – V ^ 2) / 2 * g.

hmax = V0 ^ 2/2 * g.

hmax = (45 m / s) ^ 2/2 * 10 m / s2 = 101.25 m.

t = t1 + t2, where t1 is the ascent time, the descent time.

t1 = V0 / g.

t1 = 45 m / s / 10 m / s2 = 4.5 s.

hmax = g * t2 ^ 2/2.

t2 = √ (2 * hmax / g).

t2 = √ (2 * 101.25 m / 10 m / s2) = 4.5 s.

t = 4.5 s + 4.5 s = 9 s.

Answer: the maximum lifting height was hmax = 101.25 m, the flight time of the stone was t = 9 s.