The height of a cannon ball t seconds after it is fire from a position 10 meters above the ground is given by h(t)= -4.9t^2 +14t+10 (cont. below) find the velocity at time t. At time t=2 seconds? what is maximum height of the ball? what is the velocity ?

Question

721 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-28T17:53:09

Please see the explanation below

The velocity is the derivative of the position

#h(t)=-4.9t^2+14t+10#

#v(t)=h'(t)=-9.8t+14#

The velocity at time #t# is #=-9.8t+14#

When #t=2#

#v(2)=-9.8*2+14=-5.6ms^-1#

The maximum height is reached when #v(t)=0#

#=>#, #-9.8t+14=0#

#t=14/9.8=1.429s#

The maximum height is

#h_(max.)=-4.9*1.429^2+14*1.149+10=20m#

graph{(y+4.9x^2-14x-10)(y+9.8x-14)=0 [-26.67, 31.04, -6.04, 22.83]}