91Ó°ÊÓ

The initial velocity of a projectile plays a crucial role in determining its trajectory. In our exercise, the initial velocity \(\vec{v} = (7.6 \mathrm{~m}/\mathrm{s}) \hat{\mathrm{i}} + (6.1 \mathrm{~m}/\mathrm{s}) \hat{\mathrm{j}}\) describes the speed and direction of the ball when it first leaves the ground. Here, the value \(7.6 \mathrm{~m}/\mathrm{s}\) represents the horizontal speed, while \(6.1 \mathrm{~m}/\mathrm{s}\) represents the upward speed. It’s helpful to break these into horizontal and vertical components to understand their implications separately. The horizontal component remains constant throughout the motion, while the vertical component changes due to gravity.

The maximum height is the highest point the projectile reaches in its trajectory. To reach this point, the vertical velocity component must come to zero. Using the given initial upward velocity \(v_{y0} = 6.1 \mathrm{~m}/\mathrm{s}\) and knowing gravity \(g = 9.8 \mathrm{~m/s}^2\), we employ the motion equation:\[0 = (6.1)^2 - 2(9.8)(y - 9.1)\]Solving this, we find the maximum height \(y\):\[y = 9.1 + \frac{(6.1)^2}{2 \times 9.8} \approx 10.99 \mathrm{~m}\]This indicates the ball reaches a peak height of approximately 10.99 meters before descending.

The time of flight is the total time the projectile remains in the air. It includes the time to reach the maximum height and the time to fall back to the ground. Calculating the ascent time to the maximum height, we use the formula:\[t_1 = \frac{v_{y0}}{g} = \frac{6.1}{9.8} \approx 0.62 \mathrm{~s}\]Next, for the descent from the maximum height back to the ground, we use:\[y = \frac{1}{2}gt_2^2\]Therefore,\[10.99 = \frac{1}{2}(9.8)t_2^2 \Rightarrow t_2 = \sqrt{\frac{2 \times 10.99}{9.8}} \approx 1.5 \mathrm{~s}\]Adding these times together gives the total flight time:\[t_{total} = t_1 + t_2 = 0.62 + 1.5 \approx 2.12 \mathrm{~s}\]

The horizontal distance a projectile covers is its total range. This depends on the horizontal component of the initial velocity and the total time of flight. Given the constant horizontal velocity \(v_x = 7.6 \mathrm{~m/s}\) and total flight time \(t_{total} \approx 2.12 \mathrm{~s}\), we find the range with:\[d = v_{x} \times t_{total} = 7.6 \times 2.12 \approx 16.11 \mathrm{~m}\]Thus, the ball travels around 16.11 meters horizontally.

The velocity magnitude of the projectile just before it hits the ground combines its horizontal and vertical components. As the vertical component is \(v_y = -6.1 \mathrm{~m/s}\) (downwards), and the horizontal component remains \(v_x = 7.6 \mathrm{~m/s}\), we use:\[|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{(7.6)^2 + (-6.1)^2} \approx 9.8 \mathrm{~m/s}\]This represents the speed of the ball upon impact.

The direction of the projectile’s velocity just before it hits the ground is given by the angle \(\theta\) it makes with the horizontal. Considering \(v_y = -6.1 \mathrm{~m/s}\) and \(v_x = 7.6 \mathrm{~m/s}\), we calculate:\[\theta = \arctan \left( \frac{v_y}{v_x} \right) = \arctan \left( \frac{-6.1}{7.6} \right) \approx -38.6^\circ\]The negative sign indicates an angle below the horizontal, showing the direction of descent.