Questions: A football is kicked at ground level with a speed of 17.6 m / s at an angle of 43.7° to the horizontal. Part A How much later does it hit the ground? Express your answer using three significant figures and include the appropriate units. t=Value s

Solution Steps

The initial velocity of the football is given as $v_0 = 17.6 \, \text{m/s}$ at an angle of $43.7^\circ$ to the horizontal. We need to find the horizontal and vertical components of this velocity.

• The horizontal component $v_{0x}$ is given by: $$v_{0x} = v_0 \cos \theta = 17.6 \cos 43.7^\circ$$

• The vertical component $v_{0y}$ is given by: $$v_{0y} = v_0 \sin \theta = 17.6 \sin 43.7^\circ$$

The time of flight $t$ can be determined by analyzing the vertical motion. The football will hit the ground when its vertical displacement is zero. The vertical motion is described by the equation: $$y = v_{0y} t - \frac{1}{2} g t^2$$ where $y = 0$ (since it returns to ground level) and $g = 9.81 \, \text{m/s}^2$ is the acceleration due to gravity.

Setting the equation to zero: $$0 = v_{0y} t - \frac{1}{2} g t^2$$

This simplifies to: $$t(v_{0y} - \frac{1}{2} g t) = 0$$

The non-zero solution for $t$ is: $$t = \frac{2 v_{0y}}{g}$$

Substitute $v_{0y} = 17.6 \sin 43.7^\circ$ into the equation: $$t = \frac{2 (17.6 \sin 43.7^\circ)}{9.81}$$

First, calculate $v_{0y}$: $$v_{0y} = 17.6 \sin 43.7^\circ \approx 12.1867 \, \text{m/s}$$

Now, substitute $v_{0y}$ into the time of flight equation: $$t = \frac{2 \times 12.1867}{9.81} \approx 2.4845 \, \text{s}$$

Final Answer