If x and y coordinates of a projectile as a f | vedclass

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(B) The horizontal position is given by $x = 24t$. The horizontal velocity component is $v_x = \frac{dx}{dt} = 24 	ext{ m/s}$.The vertical position i

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) The horizontal position is given by $x = 24t$. The horizontal velocity component is $v_x = \frac{dx}{dt} = 24 	ext{ m/s}$.The vertical position is given by $y = 43.6t - 4.9t^2$. The vertical velocity component is $v_y = \frac{dy}{dt} = 43.6 - 9.8t$.At time $t = 2 	ext{ s}$, the vertical velocity is $v_y = 43.6 - 9.8(2) = 43.6 - 19.6 = 24 	ext{ m/s}$.The angle $	heta$ made by the projectile with the horizontal is given by $	an 	heta = \frac{v_y}{v_x}$.Substituting the values, $	an 	heta = \frac{24}{24} = 1$.Therefore, $	heta = 	an^{-1}(1) = 45^{\circ}$.

If $x$ and $y$ coordinates of a projectile as a function of time are given as $24t$ and $43.6t - 4.9t^2$, respectively, then the angle (in degrees) made by the projectile with horizontal when $t = 2 \text{ s}$ is . . . . . . .

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