An object is thrown upward at a speed of 161 feet per second by a machine...
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?
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!!
If a ball is thrown upward at 96 feet per second from a height of 12 feet, the height of the ball can be modeled by S- 12+96t seconds after the ball is thrown. How long after the ball is thrown is the height 152 feet? 16t feet, where t is the number of t takes seconds for the ball to reach th (Type an integer or a simplified fraction. Use a comma to separate answers as needed.) e height...
A ball is thrown vertically upward with an initial velocity of 48 feet per second. The distances (in feet) of the ball from the ground after t seconds is s=48-16t2. (a) At what time t will the ball strike the ground? (b) For what time t is the ball more than 32 feet above the ground?
An object is propelled upward at an initial velocity of 32 feet per second at a. initial elevation of 48 feet above the ground. The free-fall model used to describe the motion is: s(t) = -16t2 + 326 +48 for Osts 3 (t seconds from release of object). Reference: Drop the Ball Activity The average velocity between t = 0 sec and t = 2 sec A) O feet per second B) 24 feet per second C) 12 feet per...
An object is thrown upward from a height of 6 feet at a velocity of 64 feet per second. The height of the object h is a function of time t. h(t) = -16+2 +64t + 6 At what time the function h will reach its maximum value.
A projectile is thrown upward so that its distance, in feet, above the ground after t seconds is h= -16t2+608t. What is its maximum height?
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?
If an object is thrown with a velocity of v feet per second at an angle of θ with the horizontal, then its flight can be modeled by, x = (v cos θ ) t and y = v (sin θ ) t - 16 t2 + h where t is in seconds and h is the object's initial height in feet above the ground. x is the horizontal position and y is the vertical position, and - 16 t2...
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