If an object is thrown in an upward direction from the top of a building 160 ft high at an initial speed of 30 mi/h, what is its final speed when it hits the ground? (Disregard wind resistance. Do not reflect negative direction in your answer.)
ft/s
110 fus
100 ft/s
097 ft/s
91 ft/s
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We can use  2 g H = v2^2 - v1^2    or

v2^2 = 2 g H + v1^2

Since 88 ft/sec = 60mph   we have 30 mph = 44 ft/sec

The object will return with the same speed that it had initially so the object

starts out with a downward speed of 44 ft/sec

Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2

v2^2 = (2 * 32 * 160 + 44^2) ft^2 / sec^2 = 12180 ft^2/sec^2

v2 = 110 ft/sec

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft