77. We stand 200 meters from the launch site of a balloon. The balloon rises at

Question

77. We stand 200 meters from the launch site of a balloon. The balloon rises at a constant rate of 4 meters per second. How fast is the angle of elevation of the balloon increasing 30 seconds after the launch? in) 200 m Figure 8: 0 is the angle of elevation 78. The following figure shows the power consumption in San Francisco for a day in September. P is measured in megawatts; t is measured in hours starting at midnight. Estimate the energy used on that day. Remember that power is the rate of change of energy, that is P(t) E(t). 800 600 400 200 0 Pacific Gas & Electric Figure 9: Power Consumption in a City

Explanation / Answer

77. Let b represents the distance which is 200 m.

THe balloon rises at a constant rate of 4 meter per second . So after 30 secs, height of balloon will be 4*30 =120 m

Tan theta = height / base= 120/200

theta = tan-1(120/200) =30.96 degree

tan theta = h/b

differentiating

sec2 theta (d theta/dt) = (1/b)(dh/dt)

sec230.96(d theta/dt) =(1/200)*4

d theta/dt = (1/50)(1/sec230.96) = 0.0147 degree/sec

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