A rocket runs out of fuel in the position shown and continues in unpowered flight above the atmosphere. If its velocity in this position was 567 mi/hr, calculate the maximum additional altitude hacquired and the corresponding time t to reach it. The gravitational acceleration during this phase of its flight is 30.0 ft/sec². Vertical 22% Answers: h= t= i i v = 567 mi/hr mi sec

The tank is to be placed on a hill that is 1 km high above the ground. The enemy’s base is located at a horizontal distance of 10 km away and it is in a valley 0.5 km below the ground. Air resistance is neglected. Gravitational acceleration is constant and equal to 9.8?/?2 downwards. If the bomb is to be fired perfectly horizontally, what should the launching speed of the bomb be in order to hit the target? What is the duration it would take the bomb to hit the target? Find x(t) and y(t). Consider the initial location of the tank to be (0, 1 km)? Draw the trajectory of the bomb (x and y coordinates). Indicate clearly the value of x (horizontal displacement) when the bomb reaches the ground level. 5.     Draw the (?), ??(?), and ??(?) vertical above each other. Indicate key values on the graphs such as the start, the end, the maximum, the minimum and the x and y intercepts. PART 4 and 5 PLEASE.

a.) What is the angle, measured relative to the positive x-axis, of the load's acceleration? (The correct answer is: -10.219 degrees) Just need the steps to get there  b.) Three horses pull and guide a 132 kg payload down a river, exerting the forces shown in the figure, with F1 = 29 N, F2 = 38 N, F3 = 58 N, θ1 = 30°, and θ2 = 60°. What is the magnitude of the payload's acceleration? (The correct answer is: 0.7861 m/s^2)Just need the correct steps to get there

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