We have the equation of motion $s = ut + \frac{1}{2} at^2$ , where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.

Here s = 300 m, u = 0 m/s, a = 9.81 $m/s^2$

Substituting

$300 = 0*t+\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2 = 300\\ \\ t =7.82 seconds$

Now we have v = u+at, where v is the final velocity

Here u = 0 m/s, a= 9.81 $m/s^2$ and t = 7.82 seconds

Substituting

v = 0+9.8*7.82 = 76.68 m/s

The speed with which the penny strikes the ground = 76.68 m/s.

3 0

3 years ago

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8m/s2 . The accel

Zanzabum

Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.

Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m

Now you need the final speed to use it as initial speed of the next part.

Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s

Part B) Free fall

Maximum height, y max ==> Vf = 0

Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s

ymax = yo + Vo*t - g[t^2] / 2

ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m

Answer: ymax = 10084.2m

8 0

3 years ago