Question

maximum range of projectile is 10 cm what will be its maximum range if its velocity of projection is doubled

Answer 1

Range = u²sin2theta/g
if u is doubled then,
range =(2u)²sin2theta/g=4u²sin2theta/g
Therefore the range is quadrupled.
Hence, the range is 4×10=40 cm

Answer 2

Thus the maximum range will be 40 cm.

Explanation:

We are given that:

Range of projectile = 10 cm

Now we know that;

Range = u²sin2 θ / g

If u is doubled then,

range = (2u)²sin2 θ / g = 4u²sin2 θ / g  

Therefore the range is quadrupled.

Hence, the range is 4 × 10 = 40 cm

Thus the maximum range will be 40 cm.