Horizontal projection

Classical Mechanics Level 3

A stone is thrown at a speed of 19.6 m/sec at an angle 30 degrees above the horizontal from a tower of height 490 meter. Find the time during which the stone will be in air.


The answer is 11.05.

Swaraj Shandilya by Swaraj Shandilya , 21, India

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2 solutions

We can use the formula y = V o y t − 1 2 g t 2 y=V_{oy} t-\dfrac{1}{2}g t^2 where y = − 490 y=-490 , V o y = sin ⁡ 3 0 ∘ ( 19.6 ) V_{oy}=\sin 30^\circ(19.6) and g = 9.81 g=9.81 .

Substitute:

− 490 = sin ⁡ 30 ( 19.6 ) t − 1 2 ( 9.81 ) t 2 -490=\sin 30 (19.6)t - \dfrac{1}{2}(9.81)t^2

− 490 = 9.8 t − 4.905 t 2 -490=9.8t-4.905t^2

4.905 t 2 − 9.8 t − 490 4.905t^2-9.8t-490

Use the quadratic formula to solve for t t .

t = − b ± b 2 − 4 a c 2 a = 9.8 ± ( − 9.8 ) 2 − 4 ( 4.905 ) ( − 490 ) 2 ( 4.905 ) = 980 981 ± 9709.84 9.81 ≈ 11.043682 seconds t=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}=\dfrac{9.8 \pm \sqrt{(-9.8)^2-4(4.905)(-490)}}{2(4.905)}=\dfrac{980}{981}\pm \dfrac{\sqrt{9709.84}}{9.81} \approx \boxed{\color{#69047E}11.043682~\text{seconds}}

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Swaraj Shandilya
Jun 22, 2015

Vertical motion (downward direction negative) :

Initial vertical velocity y = 19.6 sin 30o

Acceleration a = g = -9.8 m/s2

Vertical distance covered = h = 490 m

Using, h = ut + 1/2gt2

We have, 490 = - 9.8t + (1/2) 9.8t2

100 = - 2t + t2 or t2 - 2t - 100 = 0

t = \frac{2\pm \sqrt{(2^{2}-4\times 1\times \left [ -100 \right ])}}{2\times 1} =1+\sqrt{101}

t = 11.05 sec

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I've done the LaTeX job for you. So you can correct your math typing by copy-pasting my codes!

Vertical motion (downward direction negative) : Initial vertical velocity y = 19.6 s i n 3 0 0 y = 19.6 sin 30^0 Acceleration a = g = − 9.8 m / s 2 a = g = -9.8 m/s^2 Vertical distance covered = h = 490 m = h = 490 m Using, h = u t + 1 2 g t 2 h = ut + \frac{1}{2} gt^2 We have, 490 = − 9.8 t + 1 2 9.8 × t 2 490 = - 9.8t + \frac{1}{2} 9.8 \times t^2 100 = − 2 t + t 2 100 = - 2t + t^2 o r , t 2 − 2 t − 100 = 0 or, t^2 - 2t - 100 = 0 t = 2 ± ( 2 2 − 4 × 1 × [ − 100 ] ) 2 × 1 = 1 + 101 t = \frac{2 \pm \sqrt{(2^{2}-4\times 1\times \left [ -100 \right ])}}{2\times 1} =1+\sqrt{101} t = 11.05 s e c t = 11.05 sec

Muhammad Arifur Rahman - 5 years, 11 months ago

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