Question

a student can throw a ball vertically to a maximum height of 40 m .the same student can throw the ball in horizontal direction to a maximum distance of .

Answer 1

Heya.......!!!

___________________________

According to the question , a student can throw a ball vertically to a maximum height 40 m.

We have to take out the maximum distance in horizontal direction .


Firstly the vertical height which is 40 m is given by the formulae :

=> u^2 / 2g = 40

We know that Horizontal range is given by the formulae :

=> u^2 sin2θ / g

For maximum range θ = 45°

putting θ = 45° , we get sin2θ = 1

that is => u^2 / g

Now if see carefully the formulae of vertical height i.e = u^2 / 2g = 40

from this we can take out the value of u^2 /g


→ u^2 / g = 80 m


♦ Maximum horizontal range = 80 m



=================================


Hope It Helps You ☺

Answer 2

Answer:

Explanation:

Refer the below attachment.