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