Problem 1.1 We have had too many gravity problems in which a ball or stone is th

Question

Problem 1.1 We have had too many gravity problems in which a ball or stone is thrown. It is time for something more exciting. A spy refuses to answer questions and is thrown out of an airplane 1500 m above the ground. Unbeknownst to his captors, he has parachutes in the soles of his shoes. If the parachutes must be opened at a minimum of 250 m above the ground, what is the maximum amount of time the spy can wait before opening his parachutes? Assume that his initial velocity in the vertical direction is zero. Neglect air resistance and motion in the horizontal direction. Problem 1.2 The Bypss that runs next to is approximately 39.0 m wide at the intersection withYou are driving at 25.0 miles per hour the light turns yellow just as you reach the edge of the intersection, and the yellow light lasts 2.50 seconds, how fast must you accelerate to cross the intersection before the light turns red? Ignore reaction time (that is, the time it takes you to recognize that the light has changed and accelerate around 0.2 s for most people). If Problem 1.3 In Problem 1.2, we neglected the length of the car. This is usually a safe approximation since cars are only a few meters long. Also, there is usually a slight delay between the light turning red in one direction and the light turning green in the other, so that vehicles in the intersection when the light turns red are not run over. If you were driving a semi-trailer truck 18.0 m long, how fast would you have to accelerate to clear the intersection? How does this compare to Problem 1.2? All variables are as in problem 1.2. Note: If you solve 1.2 algebraically and only plug in numbers at the last step, you will already have done almost all the work for this problem, and will have to show almost no work

Explanation / Answer

Problem 1.1

As mentioned in question, we need to consider only gravitation force (acceleration due to gravity) for this motion. Initial speed in vertical direction is zero.

We need to concentrate on vertical motion only.

Distance covered in vertical direction in time t is

s= u*t + (1/2) * g * t2 ; u = 0

s = (1/2) * g * t2

t2 = 2 *s /g

t = Square root of (2 *s /g)

As we need to find maximum time to reach from 1500m to 250m (time to cover 1500-250 =1250m)

t = Square root of (2 * 1250 / 9.8) = 15.97 second

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