A small ball is thrown at an angle 45^{circ} | vedclass

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Question

(B) Initial velocity $u = 2\sqrt{2} \ m/s$ at angle $	heta = 45^{\circ}$.Horizontal component $u_x = u \cos 45^{\circ} = 2\sqrt{2} 	imes \frac{1}{\s

Vedclass · Accepted Answer

$8.2$

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(B) Initial velocity $u = 2\sqrt{2} \ m/s$ at angle $	heta = 45^{\circ}$.Horizontal component $u_x = u \cos 45^{\circ} = 2\sqrt{2} 	imes \frac{1}{\sqrt{2}} = 2 \ m/s$.Vertical component $u_y = u \sin 45^{\circ} = 2\sqrt{2} 	imes \frac{1}{\sqrt{2}} = 2 \ m/s$.Horizontal displacement in $t = 2 \ s$ is $x = u_x 	imes t = 2 	imes 2 = 4 \ m$.Vertical displacement in $t = 2 \ s$ is $y = u_y t - \frac{1}{2}gt^2 = 2(2) - \frac{1}{2}(10)(2^2) = 4 - 20 = -16 \ m$.Total displacement $S = \sqrt{x^2 + y^2} = \sqrt{4^2 + (-16)^2} = \sqrt{16 + 256} = \sqrt{272} \approx 16.49 \ m$.Mean velocity $v_{avg} = \frac{S}{t} = \frac{16.49}{2} = 8.245 \ m/s \approx 8.2 \ m/s$.

$A$ small ball is thrown at an angle $45^{\circ}$ to the horizontal with an initial velocity of $2 \sqrt{2} \ m/s$. The magnitude of the mean velocity averaged over the first $2 \ s$ is (take acceleration due to gravity $g = 10 \ m/s^2$). (in $m/s$)

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