Suppose you have thrown a ball upwards from the rooftop of IUT Academic Building 1 with a velocity of 4 m/s 4 \mathrm{~m} / \mathrm{s} at an angle of 60∘ 60^{\circ} with the horizontal line. The ball reaches the ground 10 meters away from the building. The height of the building is-

Question

Suppose you have thrown a ball upwards from the rooftop of IUT Academic Building 1 with a velocity of 4 m/s 4 \mathrm{~m} / \mathrm{s} at an angle of 60∘ 60^{\circ} with the horizontal line. The ball reaches the ground 10 meters away from the building. The height of the building is-

Accepted Answer

খ. 105.18 m

ব্যাখ্যা: প্রদত্ত:
নিক্ষেপণ বেগ $u = 4	ext{ m/s}$
নিক্ষেপণ কোণ $	heta = 60^\circ$
আনুভূমিক দূরত্ব $x = 10	ext{ m}$
অনুভূমিক বেগ ধ্রুবক: $u_x = u \cos 60^\circ = 4 	imes 0.5 = 2	ext{ m/s}$
পতনের সময়: $t = \frac{x}{u_x} = \frac{10}{2} = 5	ext{ s}$
উল্লম্ব সরণ: $y = (u \sin 60^\circ) t - \frac{1}{2}gt^2$
এখানে $g = 9.8	ext{ m/s}^2$ (বা $9.81	ext{ m/s}^2$):
$y = \left(4 	imes \frac{\sqrt{3}}{2}ight) 	imes 5 - \frac{1}{2} 	imes 9.8 	imes (5)^2 = 10\sqrt{3} - 4.9 	imes 25 = 17.3205 - 122.5 = -105.18	ext{ m}$
সুতরাং ভবনের উচ্চতা $h = |y| = 105.18	ext{ m}$।

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