How many seconds does it take for an object thrown from the ground at a speed of v=50m/s and an 30^circ angle to -4 remain in the air? (g=10m/s^2) 3 la 4 (b) 6 6 ic s s (d 2 d

To find the time of flight for an object thrown at an angle, we can use the formula:<br /><br />\[ T = \frac{2v \sin(\theta)}{g} \]<br /><br />where:<br />- \( T \) is the time of flight,<br />- \( v \) is the initial velocity,<br />- \( \theta \) is the angle of projection, and<br />- \( g \) is the acceleration due to gravity.<br /><br />Given:<br />- \( v = 50 \, \text{m/s} \)<br />- \( \theta = 30^\circ \)<br />- \( g = 10 \, \text{m/s}^2 \)<br /><br />First, calculate the sine of the angle:<br /><br />\[ \sin(30^\circ) = \frac{1}{2} \]<br /><br />Now, substitute the values into the formula:<br /><br />\[ T = \frac{2 \times 50 \times \frac{1}{2}}{10} \]<br /><br />Simplify the expression:<br /><br />\[ T = \frac{50 \times \frac{1}{2}}{10} \]<br />\[ T = \frac{25}{10} \]<br />\[ T = 2.5 \, \text{seconds} \]<br /><br />Therefore, the object remains in the air for 2.5 seconds.