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An object is thrown upward from a height of 6 feet at a velocity of 64...
a ball is thrown upward at 64 feet per second from the top of an 80 feet high building. The height of the ball can be modeled by S(t) = -16t^2 + 64t + 80(feet), where t is the number of seconds after the ball is thrown. describe the graph model
If an object is thrown upward with an initial velocity of 64 ft/sec, its height after t sec is given by h = 64t -16t^2. Find the number of seconds the object is in the air before it hits the ground.
An object is thrown upward at a speed of 161 feet per second by a machine from a height of 8 feet off the ground. The height h of the object after t seconds can be found using the equation h(t) = 16t2 + 161t + 8 When will the height be 67 feet? (Write both results, using a comma between). Round answers to the nearest hundredth. Select an answer Select an answer ict reach the ground? seconds 1 answer...
An object is thrown upward at a speed of 194 feet per second by a machine from a height of 6 feet off the ground. The height hh of the object after tt seconds can be found using the equation h(t)=−16t2+194t+6h(t)=-16t2+194t+6 When will the height be 522 feet? (Write both results, using a comma between). Round answers to the nearest hundredth. When will the object reach the ground?
(6) A ball is thrown directly upward from a height of 11 feet with an initial velocity of 192 ft/s.The function (1) =-161° +1921 +11 gives the height of the ball in feet, 7 seconds after it has been thrown. Determine the time at which the ball reaches the maximum height and find the maximum height
show clear work plz A ball is thrown upward from a height of 1000 feet above the ground, with an initial velocity of 80 feet per second. From physics it is known that the acceleration at time t is a(t) = -32 feet per second squared. Use antiderivatives to find s(t), the function giving the height of the ball at timet.
An object is thrown upward at a speed of 182 feet per second by a machine from a height of 14 feet off the ground. The height hh of the object after tt seconds can be found using the equation h=−16t^2+182t+14h=- When will the height be 515 feet? Round to two place When will the object reach the ground? Round to two places. USE EXCEL SOLVER!!
A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function h(t)=104t-16t^2 .After how long will it reach its maximum height? Do not round your answer.
if an object is thrown vertically upward with an initial velocity of v, from an original position of s, the height h at any time t is given by: h = -16t^2 + vt + s If an object is thrown vertically upward with an initial velocity of v, from an original position of s, the height h at any time t is given by h16t2 +vt+s (where h and s are in ft, t is in seconds and v...
A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s(t)=96t-16t^2a.) at what time will the ball strike the groundb.) for what time t is the ball more than 128 feet above the ground?c.) when will the ball reach its highest peak? how high is it above the ground?