Question

A man throws a packet from a tower directly aiming at his friend who is standing at a certain distance from the base which is same as a height of the tower. If packet is thrown with a speed of 4 m/s and it hits the ground midway between the tower base and his friend then height of the tower is (g=10 m/s2)

Answer 1

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Answer 2

Answer:

3.2m

Explanation:

As it is a projection at an angle ∅ below horizontal

h=usin∅ T+$\frac{1}{2}$g$T^{2}$

where T is the time of flight

this can be written as $\frac{1}{2}$g$T^{2}$+usin∅ T-h=0

now using the quadractic formula T=(-usin∅±$\sqrt{usin∅^{2} -4(1)(\frac{1}{2}*g) }$)/$2*\frac{1}{2} g$

on substituting values we get $\frac{\frac{-4}{\sqrt{2}} ±\sqrt{8+20x} }{10}$ (ignore the A in eq its some bug)

we get $\frac{4}{\sqrt{2} }$ because as he is throwing directly to his friend and the height of the tower and the distance between him and the tower is equal

therefore ∅ =45 I have assumed height as x

as T is +ve reject -ve

Substituting T in eq R=ucos∅T(As it is a projection at an angle ∅ below horizontal)

(note:r=x/2(given))

$\frac{x}{2} =\frac{4}{\sqrt{2} } *(\frac{-4}{\sqrt{2} }+\sqrt{8+20x})/10$

$\frac{x}{2} =\frac{-4}{5} +\frac{\sqrt{16+40x}}{5}$

$\frac{5x}{2} +4=\sqrt{16+40x}$

on squaring both sides

$\frac{25x^{2} +64+80x}{4} =16+40x$

$25x^{2} +64+80x =64+160x$

$25x^{2} =80x$

$x=\frac{80}{25}$

=3.2m

hope it helps