Answer to Question #214073 in Physics for Vincent

Question

A body is projected vertically upwards from ground with an initial velocity of 400m/s.

Calculate:

a)the time taken to reach the ground

b)the velocity when the body is 1.9km above ground level

c)the time taken to reach a vertical hight of 2.6km

Answer 1

(a) Let's first find the time taken to reach the maximum height:

$$v=v_0-gt,$$ $$0=v_0-gt,$$ $$t=\dfrac{v_0}{g}=\dfrac{400\ \dfrac{m}{s}}{9.8\ \dfrac{m}{s^2}}=40.82\ s.$$

Finally, we can find the total time taken to reach the ground:

$$t_{tot}=2t=2\cdot40.82\ s=81.64\ s.$$

(b) Let's first find the time taken to reach the height of 1.9 km above ground level:

$$y=v_0t-\dfrac{1}{2}gt^2,$$ $$1900=400t-4.9t^2,$$ $$4.9t^2-400t+1900=0.$$

This quadratic equation has two roots: $t_1=76.57\ s$ and $t_2=5.06\ s.$ The second root corresponds to the case when the body rises up and reaches the height of 1.9 km above ground level first time (the first one corresponds to the case when the body begins to fall down and reaches the height of 1.9km once again). Therefore, the correct answer is $t=5.06\ s$.

Finally, we can find the velocity when the body is 1.9 km above ground level:

$$v=v_0-gt=400\ \dfrac{m}{s}-9.8\ \dfrac{m}{s^2}\cdot5.06\ s=350.4\ \dfrac{m}{s}.$$

(c) We can find the time taken to reach a vertical height of 2.6 km from the kinematic equation:

$$y=v_0t-\dfrac{1}{2}gt^2,$$ $$2600=400t-4.9t^2,$$ $$4.9t^2-400t+2600=0.$$

This quadratic equation has two roots: $t_1=74.51\ s$ and $t_2=7.12\ s.$ The second root corresponds to the case when the body rises up and reaches the height of 2.6 km above ground level first time (the first one corresponds to the case when the body begins to fall down and reaches the height of 2.6km once again). Therefore, the correct answer is $t=7.12\ s$.