an object is launched at a speed of 20.0 m/s from the top of a tall tower. The h

Question

an object is launched at a speed of 20.0 m/s from the top of a tall tower. The height y of the object as a function of time is given by y(t)=-4.9t^2+19.32t+60. find the horizontal distance traveled by the object before it hits the ground an object is launched at a speed of 20.0 m/s from the top of a tall tower. The height y of the object as a function of time is given by y(t)=-4.9t^2+19.32t+60. find the horizontal distance traveled by the object before it hits the ground an object is launched at a speed of 20.0 m/s from the top of a tall tower. The height y of the object as a function of time is given by y(t)=-4.9t^2+19.32t+60. find the horizontal distance traveled by the object before it hits the ground

Explanation / Answer

y(t)=-4.9t^2+19.32t+60

at t = 0

y = height of tower = h = 60 m

when it hits the ground y =0

0 = -4.9t^2 + 19.32t +60

4.9t^2 - 19.32t - 60 = 0

solving the equation

t = 5.987 s and t = -2.045 s

negative can be ignored

so t = 5.987 s (total time of flight)

when y(t) = maximum dy/dt = 0

d/dt ( -4.9t^2+19.32t+60) = 0

-9.8t + 19.32 = 0

t = 19.32/9.8 = 1.9714 s

after 1.9714 s the object is at its highest position

t = u*sin(theta)/g

1.9714 = 20*sin(theta)/9.8

sin(theta) = 0.966

theta = 75.016 degrees

R = ucos(theta)*total time of flight

R = 20*cos(75.016)*5.9878 = 30.962 m

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