Question

A projectile thrown at 45degree just cross the top of pole whose foot is 6 m from point of projection. The projectile falls on the other side at a distance 3m from foot of pole. Height of pole is????
(a) 1.5 m
(b) 2 m
(c) 2.5 m
(d) 3 m

Answer 1

Answer on Question #40826, Physics, Mechanics

A projectile thrown at $45{}^{\circ}$ just cross the top of pole whose foot is $6\mathrm{m}$ from point of projection. The projectile falls on the other side at a distance $3\mathrm{m}$ from foot of pole. Height of pole is?

(a) $1.5\mathrm{m}$

(b) $2\mathrm{m}$

(c) $2.5\mathrm{m}$

(d) $3\mathrm{m}$

Solution:

Given:

$x_{1} = 6\mathrm{m}$

$x_{2} = 3\mathrm{m}$

$\theta = 45{}^{\circ}$

$h = ?$

Neglecting air resistance, the projectile is subject to a constant acceleration $g = 9.81 \, \text{m/s}^2$ , due to gravity, which is directed vertically downwards.

Equations related to trajectory motion (projectile motion) are given by

Horizontal distance, $x = v_{0x}t$

Vertical distance, $y = v_{0y}t - \frac{1}{2} gt^2$

Horizontal range, $R = \frac{v_0^2\sin 2\theta}{g}$

where $v_{0}$ is the initial velocity.

We have

$$R = 6 + 3 = 9 \mathrm {m}$$

Thus,

$$v _ {0} = \sqrt {\frac {R g}{\sin 2 \theta}} = \sqrt {\frac {9 \cdot 9 . 8 1}{1}} = 9. 4 \mathrm {m / s}$$

Time to reach a pole is

$$t _ {1} = \frac {x _ {1}}{v _ {0 x}} = \frac {x _ {1}}{v _ {0} \cos \theta} = \frac {6}{9 . 4 \cdot \cos 4 5 {}^ {\circ}} = 0. 9 0 2 7 \mathrm {s}$$

Vertical distance,

$$h = y = v _ {0} \sin \theta t _ {1} - \frac {1}{2} g t _ {1} ^ {2} = 9.4 \cdot \sin 45{}^{\circ} \cdot 0.9027 - \frac {9.81 \cdot 0.9027^2}{2} = 2 \mathrm{~m}$$

Answer. (b) 2 m.