Question

A projectile shot at an angle 60 deg to the horizontal , strikes a wall 30 m away at a point 15 m above the point of projection . (g=10)
(i) find the speed of projection.
(ii) find the magnitude of velocity of the projectile when it strikes the wall.

Answer 1

GiVeN :-

A projectile shot at an angle 60° to the horizontal, strikes a wall 30 m away at a point 15 m above the point of projection.

• x = 30 m
• y = 15 m
• θ = 60°
• g = 10 m/s²

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To DeTeRmInE :-

(i) The speed of projection.

(ii) The magnitude of velocity of the projectile when it strikes the wall.

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SoLuTiOn :-

(i) We know,

$\displaystyle{\tt{x=\frac{u}{2}t }}$

$\displaystyle{\tt{\hookrightarrow 30=\frac{u}{2}t }}\\\\\displaystyle{\tt{\hookrightarrow t=\frac{60}{u} }}\:\:\:\: \cdots \tt{(i)}$

Vertical component $\sf{u_y}$ = usin60°

$\sf{\implies u_y=u\times\dfrac{\sqrt{3} }{2} }$

We also know,

$\displaystyle{\tt{s=u_yt-\frac{1}{2}gt^2 }}\\\\\displaystyle{\tt{\hookrightarrow 15=u\frac{\sqrt{3}}{2}t-\frac{1}{2}\times10\times t^2 }}\\\\\displaystyle{\tt{\hookrightarrow 15=u\frac{\sqrt{3}}{2}\times \frac{60}{u}-5 \times (\frac{60}{u} )^2 }}\:\:\:\: \cdots \tt{[From\ (i)]}\\\\\displaystyle{\tt{\hookrightarrow 15=\frac{60\sqrt{3}u}{2u}-\frac{18000}{u^2} }}\\\\\displaystyle{\tt{\hookrightarrow 15=\frac{30\sqrt{3}}{1}-\frac{18000}{u^2} }}$

$\displaystyle{\tt{\hookrightarrow 15=\frac{30\sqrt{3}u^2-18000}{u^2} }}\\\\\displaystyle{\tt{\hookrightarrow 15u^2=30\sqrt{3}u^2-18000}}\\\\\tt{(Taking\ \sqrt{3}=1.7)}\\\\\displaystyle{\tt{\hookrightarrow 15u^2=51u^2-18000}}\\\\\displaystyle{\tt{\hookrightarrow 51u^2-15u^2-18000=0}}\\\\\displaystyle{\tt{\hookrightarrow 36u^2=18000}}\\\\\displaystyle{\tt{\hookrightarrow u^2=\frac{18000}{36} }}\\\\\displaystyle{\tt{\hookrightarrow u^2=500}}\\\\\displaystyle{\tt{\hookrightarrow u=\sqrt{500}}}$

$\boxed{\displaystyle{\tt{\hookrightarrow u=22.36\ m/s}}}$

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(ii) Magnitude of the velocity when it hit the wall can be calculated as

v² = u² - 2gh

$\displaystyle{\tt{\hookrightarrow v^2=(22.36)^2-2\times10\times15}}\\\\\displaystyle{\tt{\hookrightarrow v^2=499.9696-300}}\\\\\displaystyle{\tt{\hookrightarrow v^2=199.9696}}\\\\\displaystyle{\tt{\hookrightarrow v=\sqrt{199.9696} }}\\\\\boxed{\displaystyle{\tt{\hookrightarrow v=14.14\ m/s}}}$