Question

Find the speed of a rock which is thrown off the top of a 20 m tall building at 15 m/s when it makes contact with a bird which is flying at an altitude of 10 m above the ground.

Answer 1

Answer:

$v_f=20.52\frac{m}{s}$

Explanation:

In this case the rock is under an uniformly accelerated motion. Thus, we use the kinematic equation that relates the final speed of an object with its initial speed, its acceleration and its traveled distance.

$v_f^2=v_0^2+2a\Delta y\\v_f^2=v_0^2+2a(y_2-y_1)\\v_f^2=(15\frac{m}{s})^2+2(9.8\frac{m}{s^2})(20m-10m)\\v_f=\sqrt{421\frac{m^2}{s^2}}\\v_f=20.52\frac{m}{s}$

Answer 2

Answer:

20.62m/s

Explanation:

Using one of the equations of motion given by;

v² = u² + 2as     -------------------------(i)

where;

v = final velocity of the object (rock)

u = initial velocity of the object (rock) = 15m/s

a = acceleration due to gravity of the rock = g = +10m/s² (since the direction of the rock is downwards in the direction of gravity towards the flying bird).

s = distance traveled by the rock.

Notice that the rock is thrown from a 20m height to hit a bird flying at an altitude of 10m. This means that the distance (s) travelled by the rock is;

20m - 10m = 10m

Substituting the values of u, a and s into equation (i) gives;

v² = 15² + (2 x 10 x 10)

v² = 225 + 200

v² = 425

v = √425

v = 20.62m/s

Therefore the speed of the rock when it makes contact with the bird is 20.62m/s