Question

from top of a building 125 m tall a ball is projected horizontally with a speed of 120 m/s find when and with what speed dose it strike the ground

Answer 1

Answer:

130m/s

Explanation:

u=120m/s

h=125m

and g=10m/s^2

so applying third equation of motion,

v^2=u^2+2gh

= 120^2 +2×10×125

=14400+2500

=16900

or, v=130m/s

Answer 2

Answer:

Speed, |V| = 130 m/s

Explanation:

Given;-

        h = 125 m ( height of the building )

         v = 120 m/s ( speed of horizontal projection )

Since,

 Speed, |V| = √V²x + V²y

Now, Time of Flight, Tf = √2h/ g  [where h is the height of the building & g is acceleration]

Then,       Tf = √2h/g

                Tf = √2 × 125 / 10

                Tf = √250/ 10

                Tf = 5 s

Now, finding components if V;

                Vx = 120 m/s ( remains constant )

Also,         Vy = gt

                Vy = 10 × 5

                Vy = 50 m/s    

Now, finding speed with which the ball strikes the ground;

                |V| = √V²x + V²y  

                |V| = √50² + 120²

                |V| = 130 m/s

Therefore, the speed with which the ball strikes the ground is 130 m/s .