Question

*. A particle is projected from the ground with
some initial velocity making an angle of 45°
with the horizontal. It reaches a height of
7.5 m above the ground while it travels a
horizontal distance of 10 m from the point of
projection. Find the initial speed of projection.

Answer 1

Answer:

Here in this case,

\theta =45^o, g=10 m/s^2θ=45

o

,g=10m/s

2

, Horizontal distance (x)=10m(x)=10m,

Vertical distance (y)=7.5m(y)=7.5m

y=(\tan\theta)xy=(tanθ)x -\left(\dfrac{g}{2u^2\cos^2\theta}\right)x^2−(

2u

2

cos

2

θ

g

​

)x

2

\Rightarrow 7.5=(1\times 10)-\left(\dfrac{10}{2u^2\times \dfrac{1}{2}}\right)\times 100⇒7.5=(1×10)−

⎝

⎜

⎜

⎛

​

2u

2

×

2

1

​

10

​

⎠

⎟

⎟

⎞

​

×100

\Rightarrow \dfrac{1000}{u^2}=2.5⇒

u

2

1000

​

=2.5

\Rightarrow u^2=\dfrac{1000}{2.5}=400⇒u

2

=

2.5

1000

​

=400

\Rightarrow u=\sqrt{400}=20⇒u=

400

​

=20 m/s