A ball is thrown vertically upward from the ground with a speed of 80 ft/sec. If g= 32 ft/sec², then the ball will be at a height of 96 feet above the ground after time (a) 2 and 3 sec (b) 3 sec only (c) 2 sec only (d) 1 and 2 sec
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![A ball is thrown vertically upward from the ground with a speed of 80 ft/sec. If g = 32 ft/sec², then the
ball will be at a height of 96 feet above the ground after time
(a) 2 and 3 sec
78.
(b) 3 sec only
(c) 2 sec only
(d) 1 and 2 sec](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7dd65b9d-f5ad-43c7-9488-e4ecddab8b86%2F91a12517-7b5b-4baa-a081-da7a0cf0ca80%2Feksq4ev_processed.png&w=3840&q=75)
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- Here are three vectors in meters: d1 = -6.60î + 8.70ĵ + 3.50k d 2 = - 2.00î – 4.00ĵ + 2.00k d 3 = 2.00î + 3.00ĵ + 1.00k What results from (a) đ1· ( đ 2 + đ 3 ), (b) đ 1 · ( đ 2 × đ 3), and d 1 x ( d 2 + d 3 ) (c), (d) and (e) for i,j and k components respectively)? (a) Number i Units (b) Number Units (c) Number i Units (d) Number Units (e) Number i UnitsYou jump off the end of a ski jump. Your height in meters relative to the height of the ski jump after t seconds is given by h(t)=-5t^(2)+12t. How high will you be compared to the end of the ski jump after 6 seconds?A cat chases a mouse across a 1.7 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 1.5 m from the edge of the table. The acceleration of gravity is 9.81 m/s 2 . What was the cat’s speed when it slid off the table? Answer in units of m/s.
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- A ball of 5kg is shot vertically upward with 19.6m/s. What is the maximum height it can reach (in meter)?The equation r(t) = ( sin t)i + ( cos t)j + (t) k is the position of a particle in space at time t. Find the particle's velocity and acceleration vectors. π Then write the particle's velocity at t= as a product of its speed and direction. The velocity vector is v(t) = (i+j+ k.A group of physics students hypothesize that for an experiment they are performing, the speed of an object sliding down an inclined plane will be given by the expression v=2gd(sin(θ)−μkcos(θ))−−−−−−−−−−−−−−−−−−−√v=2gd(sin(θ)−μkcos(θ)). For their experiment, d=0.725meterd=0.725meter, θ=45.0∘θ=45.0∘, μk=0.120μk=0.120, and g=9.80meter/second2g=9.80meter/second2. Use your calculator to obtain the value that their hypothesis predicts for vv.
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