This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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4.1 Define the term projectile. A projectile is an object upon which the only force acting is gravity. 4.2 Calculate the: 4.2.1 The value of X We will use the kinematic equation for displacement. Let's take upwards as the positive direction. Given: Initial velocity (v_i) = 17.5 m·s^-1 Time (t) = 5 s Acceleration due to gravity (a) = -9.8 m·s^-2 Step 1: Write down the appropriate kinematic equation. y = v_i t + (1)/(2) a t^2 Step 2: Substitute the known values into the equation. y = (17.5)(5) + (1)/(2)(-9.8)(5)^2 Step 3: Calculate the displacement. y = 87.5 - 4.9(25) y = 87.5 - 122.5 y = -35 m The height X is the magnitude of the displacement. X = 35 m 4.2.2 Speed V with which the ball hits the ground We will use the kinematic equation for final velocity. Given: Initial velocity (v_i) = 17.5 m·s^-1 Time (t) = 5 s Acceleration due to gravity (a) = -9.8 m·s^-2 Step 1: Write down the appropriate kinematic equation. v_f = v_i + at Step 2: Substitute the known values into the equation. v_f = 17.5 + (-9.8)(5) Step 3: Calculate the final velocity. v_f = 17.5 - 49 v_f = -31.5 m·s^-1 The speed V is the magnitude of the final velocity. V = 31.5 m·s^-1 4.3 Calculate the value of V_Y. For ball Q, it is thrown vertically upwards from the same height X. Let's take upwards as the positive direction. Given: Displacement ( y) = -X = -35 m (since it ends up 35 m below its starting point) Acceleration due to gravity (a) = -9.8 m·s^-2 Final speed = (6)/(7) V = (6)/(7) (31.5) = 27 m·s^-1 Since the ball hits the ground, its final velocity (v_f) is downwards: v_f = -27 m·s^-1 Step 1: Write down the appropriate kinematic equation. v_f^2 = v_i^2 + 2a y Step 2: Substitute the known values into the equation. (-27)^2 = (V_Y)^2 + 2(-9.8)(-35) Step 3: Simplify and solve for (V_Y)^2. 729 = (V_Y)^2 + 686 (V_Y)^2 = 729 - 686 (V_Y)^2 = 43 Step 4: Calculate V_Y. V_Y = sqrt(43) V_Y ≈ 6.56 m·s^-1 4.4 Draw a position-time graph for ball Q from the moment it was thrown upwards until it strikes the ground. Take the ground as the zero position. For ball Q: Initial position (y_i) = X = 35 m (since the ground is the zero position) Initial velocity (v_iQ) = V_Y ≈ 6.56 m·s^-1 Acceleration (a) = -9.8 m·s^-2 Key points for the graph: Initial point: At t=0 s, y = 35 m. So, (0, 35). Time to reach maximum height (t_peak): At maximum height, v_f = 0. v_f = v_iQ + at_peak 0 = 6.56 + (-9.8)t_peak t_peak = (6.56)/(9.8) ≈ 0.67 s Maximum height (y_max): y_max = y_i + v_iQt_peak + (1)/(2)at_peak^2 y_max = 35 + (6.56)(0.67) + (1)/(2)(-9.8)(0.67)^2 y_max ≈ 35 + 4.3952 - 2.201 y_max ≈ 37.19 m So, the peak of the trajectory is at approximately (0.67 s, 37.19 m). Time to hit the ground (t_ground): At the ground, y_f = 0 m. y = y_f - y_i = 0 - 35 = -35 m y = v_iQt_ground + (1)/(2)at_ground^2 -35 = (6.56)t_ground + (1)/(2)(-9.8)t_ground^2 -35 = 6.56t_ground - 4.9t_ground^2 4.9t_ground^2 - 6.56t_ground - 35 = 0 Using the quadratic formula t = -b ± sqrt(b^2 - 4ac)2a: t_ground = -(-6.56) ± sqrt((-6.56)^2 - 4(4.9)(-35))2(4.9) t_ground = 6.56 ± sqrt(43.0336 + 686)9.8 t_ground = 6.56 ± sqrt(729.0336)9.8 t_ground = (6.56 ± 27.0006)/(9.8) Since time must be positive: t_ground = (6.56 + 27.0006)/(9.8) = (33.5606)/(9.8) ≈ 3.42 s So, the ball hits the ground at approximately (3.42 s, 0 m). The graph will be a downward-opening parabola starting at (0, 35), reaching a maximum height at approximately (0.67, 37.19), and ending at (3.42, 0). `mermaid graph TD A[Start (0, 35)] --> B[Peak (0.67, 37.19)] B --> C[End (3.42, 0)] style A fill:#fff,stroke:#333,stroke-width:2px style B fill:#fff,stroke:#333,stroke-width:2px style C fill:#fff,stroke:#333,stroke-width:2px subgraph Position-Time Graph for Ball Q direction LR t0[0] -- x-axis (Time in s) --> t1[0.67] t1 -- x-axis (Time in s) --> t2[3.42] y0[0] -- y-axis (Position in m) --> y1[35] y1 -- y-axis (Position in m) --> y2[37.19] point0(0, 35) pointPeak(0.67, 37.19) pointEnd(3.42, 0) point0 --- pointPeak pointPeak --- pointEnd end ` The graph would look like a parabola opening downwards, starting at a positive position (35 m), rising slightly to a maximum, and then curving downwards to reach 0 m at approximately 3.42 seconds. Got more? Send 'em.