Question #27767

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Answer 1 · 2017-10-11T20:29:09

See below.

The kinematic movement equations are

#(x,y) = (v_0 costheta_0 t, h+v_0 sintheta_0 t -1/2g t^2)#

calling #(x_1,0)# the point of impact after #t_1# seconds we have

#(x_1,0) = (v_0 costheta_0 t_1, h+v_0 sintheta_0 t_1 -1/2g t_1^2)#

or

#{(x_1 =v_0 costheta_0 t_1 ),(0 =h+v_0 sintheta_0 t_1 -1/2g t_1^2 ):}#

now solving for #h, v_0# we obtain

#{(v_0 = x_1/(costheta_0 t_1)), (h =(costheta_0 g t_1^2 - 2 sintheta_0 x_1)/(2 costheta_0)):}#

and then the speed just before the impact is computed using mechanical energy. So

#1/2mv_0^2+h mg = 1/2 m v_1^2# and then

#v_1 = sqrt(v_0^2+2hg)#

The item c) is left as an exercise.