A projectile launched with an angle of 45.0° above the horizontal is in the air for 7.5 s. What is the initial velocity of the projectile?

Question

Elena Albarran · Accepted Answer

I found: #v_i=51.9~~52m/s#

Have a look:

Leo Abeyta · Answer

undefined To find the initial velocity of the projectile, you can use the horizontal and vertical components of motion. Since the angle is 45 degrees, the initial vertical velocity equals the initial horizontal velocity. Use the time of flight and the vertical motion equation to find the initial vertical velocity, then use it to find the initial horizontal velocity using the horizontal motion equation.

Isabella Alvarez · Answer

undefined To find the initial velocity of the projectile, you can use the kinematic equation for projectile motion:

$$ \Delta y = v_i \cdot t \cdot \sin(\theta) - \frac{1}{2} g t^2 $$

Given:
- $\Delta y$ (vertical displacement) is 0 (assuming the projectile returns to its initial height).
- $t$ (time of flight) is 7.5 s.
- $\theta$ (launch angle) is 45.0°.
- $g$ (acceleration due to gravity) is approximately $9.81 \, \text{m/s}^2$.

Rearranging the equation to solve for $v_i$:

$$ v_i = \frac{\Delta y + \frac{1}{2} g t^2}{t \cdot \sin(\theta)} $$

Substitute the given values:

$$ v_i = \frac{0 + \frac{1}{2} \times 9.81 \times (7.5)^2}{7.5 \times \sin(45^\circ)} $$

$$ v_i = \frac{0 + \frac{1}{2} \times 9.81 \times 56.25}{7.5 \times \frac{\sqrt{2}}{2}} $$

$$ v_i = \frac{0 + 274.8375}{7.5 \times \frac{\sqrt{2}}{2}} $$

$$ v_i \approx \frac{274.8375}{7.5 \times \frac{\sqrt{2}}{2}} $$

$$ v_i \approx \frac{274.8375 \times 2}{7.5 \times \sqrt{2}} $$

$$ v_i \approx \frac{549.675}{7.5 \times \sqrt{2}} $$

$$ v_i \approx \frac{549.675}{10.6066} $$

$$ v_i \approx 51.824 \, \text{m/s} $$

So, the initial velocity of the projectile is approximately $51.824 \, \text{m/s}$.