Question

The equation of the path of the projectile is y = 0.5x – 0.04x2. The initial speed of the projectile is
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Answer 1

Answer:

y=0.5x−0.04x

2

=x(0.5−0.04x)

When y=0,x=0orR

∴R=

0.04

0.5

=12.5

y=xtanθ[1−

R

x

]⇒tanθ=

2

1

;R=12.5m

But R=

g

4u

2

sinθcosθ

⇒u=

2sinθcosθ

Rg

=

2×

5

1

×

5

2

12.5×10

=12.5m/s