This time, William Tell is shooting at an apple that hangs on a tree (see the figure below). The apple is a horizontal distance of 20.0 m away and at a height of 4.00 m above the ground, if the arrow is released from a height of 1.00 m above the ground and hits the apple 0.700 s later, what is the arrow's initial velocity. Ans.: \( 29.59 \mathrm{~m} / \mathrm{s}, 15.12^{\circ} \mathrm{N} \) of E

Given: - Horizontal distance between the apple and the point where the arrow is released: 20.0 m - Height of the apple above the ground: 4.00 m - Height of the arrow above the ground when released: 1.00 m - Time taken for the arrow to hit the apple: 0.700 s We need to find the initial velocity of the arrow. To solve this problem, we can use the equations of motion under gravity. Since the arrow is released from a height of 1.00 m and hits the apple at a height of 4.00 m, the vertical displacement is 3.00 m. The vertical displacement can be calculated using the equation: \[ s = ut + \frac{1}{2}gt^2 \] where: - \( s \) is the vertical displacement (3.00 m) - \( u \) is the initial velocity (which we need to find) - \( t \) is the time taken (0.700 s) - \( g \) is the acceleration due to gravity (approximately 9.81 m/s^2) Rearranging the equation to solve for \( u \), we have: \[ u = \frac{s - \frac{1}{2}gt^2}{t} \] Now, we can calculate the initial velocity of the arrow. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(3-0.5\times 9.81\times 0.7^{2}\right)}{0.7}\) - step1: Remove the parentheses: \(\frac{3-0.5\times 9.81\times 0.7^{2}}{0.7}\) - step2: Convert the expressions: \(\frac{3-0.5\times 9.81\left(\frac{7}{10}\right)^{2}}{0.7}\) - step3: Multiply: \(\frac{3-\frac{48069}{20000}}{0.7}\) - step4: Subtract the numbers: \(\frac{\frac{11931}{20000}}{0.7}\) - step5: Convert the expressions: \(\frac{\frac{11931}{20000}}{\frac{7}{10}}\) - step6: Multiply by the reciprocal: \(\frac{11931}{20000}\times \frac{10}{7}\) - step7: Reduce the numbers: \(\frac{11931}{2000}\times \frac{1}{7}\) - step8: Multiply the fractions: \(\frac{11931}{2000\times 7}\) - step9: Multiply: \(\frac{11931}{14000}\) The initial velocity of the arrow is approximately 0.8522 m/s. To find the horizontal component of the initial velocity, we can use the horizontal distance and the time taken: \[ v_x = \frac{d}{t} \] where: - \( v_x \) is the horizontal component of the initial velocity - \( d \) is the horizontal distance (20.0 m) - \( t \) is the time taken (0.700 s) Now, we can calculate the horizontal component of the initial velocity. Calculate the value by following steps: - step0: Calculate: \(\frac{20}{0.7}\) - step1: Convert the expressions: \(\frac{20}{\frac{7}{10}}\) - step2: Multiply by the reciprocal: \(20\times \frac{10}{7}\) - step3: Multiply: \(\frac{20\times 10}{7}\) - step4: Multiply: \(\frac{200}{7}\) The horizontal component of the initial velocity is approximately 28.57 m/s. To find the angle of the initial velocity, we can use the horizontal and vertical components: \[ \theta = \arctan\left(\frac{v_y}{v_x}\right) \] where: - \( \theta \) is the angle of the initial velocity - \( v_y \) is the vertical component of the initial velocity (0.8522 m/s) - \( v_x \) is the horizontal component of the initial velocity (28.57 m/s) Now, we can calculate the angle of the initial velocity. Calculate the value by following steps: - step0: Calculate: \(\arctan\left(\frac{0.8522}{28.57}\right)\) - step1: Divide the terms: \(\arctan\left(\frac{4261}{142850}\right)\) The angle of the initial velocity is approximately 0.02982 radians. Converting this to degrees, we have: \[ \theta \approx 0.02982 \times \frac{180}{\pi} \] Now, we can calculate the angle in degrees. Calculate the value by following steps: - step0: Calculate: \(\frac{0.02982\times 180}{\pi }\) - step1: Multiply the numbers: \(\frac{5.3676}{\pi }\) - step2: Convert the expressions: \(\frac{\frac{13419}{2500}}{\pi }\) - step3: Multiply by the reciprocal: \(\frac{13419}{2500}\times \frac{1}{\pi }\) - step4: Multiply the fractions: \(\frac{13419}{2500\pi }\) The angle of the initial velocity is approximately 1.70856 degrees. Therefore, the initial velocity of the arrow is approximately 28.57 m/s at an angle of 1.70856 degrees North of East.