Question

From a point on the ground a particle is projected with initial velocity u such that its horizontal range is maximum. Find the magnitude of average velocity during its ascent.

Answer 1

Hii dear,

# Answer- √5u/2√2

# Explaination-
For horizontal range to be maximum, angle of projection should be 45°.

Now,
R = u^2/g
H = u^2/4g
T = √2u/g

For convenience, we'll assume path to be straight line.
Using Pythagorean theorem,
D^2 = H^2 + (R/2)^2
D^2 = (u^2/4g)^2 + ^2u^2/2g)
D^2 = (5u^4)/(16g^2)
D = √5u^2/4g

Now, average velocity is given by,
v = Distance/time of ascent
v = 2D/T
v = 2×(√5u^2/4g)/(√2u/g)
v = √5u/2√2

Hence average velocity will be √5u/2√2.

Hope that is helpful...

Answer 2

Hii dear,

# Answer- √5u/2√2

# Explaination-

For horizontal range to be maximum, angle of projection should be 45°.

Now,

R = u^2/g

H = u^2/4g

T = √2u/g

For convenience, we'll assume path to be straight line.

Using Pythagorean theorem,

D^2 = H^2 + (R/2)^2

D^2 = (u^2/4g)^2 + ^2u^2/2g)

D^2 = (5u^4)/(16g^2)

D = √5u^2/4g

Now, average velocity is given by,

v = Distance/time of ascent

v = 2D/T

v = 2×(√5u^2/4g)/(√2u/g)

v = √5u/2√2

Hence average velocity will be √5u/2√2.

Hope that is helpful...