Question

A body is thrown horizontally from the top of a tower and strikes the ground after three seconds at an angle of 45∘ with the horizontal. Find the height of the tower and the speed with which the body was projected. (Takeg=9.8m/s2)

Answer 1

Given :-

▪ A body is thrown horizontally from the top of a tower and strikes the ground after 3 seconds while making an angle of 45° with the horizontal.

▪ Acceleration due to gravity, g = 9.8 m/s²

To Find :-

▪ Height of the tower.

▪ Speed at which the body was projected.

Solution :-

It is clear that the motion of the body is parabolic. Also, It is given that the body was thrown horizontally which means from the maximum height of the parabolic path. Which is also the height of the tower.

Further, the body strikes the ground at an angle of 45°.

We know,

⇒ Time of flight = 2u sin θ / g

But, We are given the time only from the maximum height to the ground. So, for the whole journey, Time of flight = 6 seconds

⇒ 6 = 2 u sin 45° / 9.8

⇒ 3×9.8 = u / √2

⇒ 29.4×2 = √2 u

⇒ 58.8 / √2 = u ...(1)

So, Let us find the maximum height or simply the height of the tower.

⇒ Height = u²sin²θ / 2g

⇒ Height = (58.8)² × 1/2 × 1/2 / 2×9.8

⇒ Height = 3457.44 × 1/4 / 19.6

⇒ Height = 44.1 m

Now, Let us find the speed at which the body was projected. which is ofcourse the magnitude of horizontal component.

⇒ Horz. comp. = u cos 45°

⇒ Horz. comp. = 58.8 / √2 × 1/√2

⇒ Horz comp. = 58.8 × 1/2

⇒ Horz. comp. = 29.4 m/s

So, We have

• Height of tower = 44.1 m
• Speed at which it was thrown = 29.4 m/s

Answer 2

$\huge\blue{Hello\:jii✌}$

Given :

$u_{y} \: \: \: \: \: \: \: \: a_{y} = g = 9.8m/s$

Let the projectile's initial velocity be u.

Time of flight T=3 s (given)

Using

$t = \sqrt{ \frac{2h}{g} } \: \: \: \ \:⟹ h \frac{gT {}^{2} }{2}$

$h = \frac{9.8 \times 3 {}^{2} }{2} = 44.1 \: m$

$x \: direction : \: a_{x} = 0 \: \: \: \: \:⟹ v _{x} = u$

$y \: direction \: : v _{y}(?) = u _{y} + a _{y}</p><p>T</p><p>$

$v _{y} = 9.8 \times 3 = 29.4m /s$

$Also \frac{v_{y} }{v_{x} } = \tan45° = 1$

$⟹ v _{x} = v_{y} = 29.4m/s$

Thus initial speed of the projectile u=

$v _{x} = 29.4 \: m/s$

That's what I found

$Hope\:it\:help$

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