Physics Motion Problem?

Question

i have this question that i can't answer please help and show working.

In a movie, Tarzan runs along level ground towards the edge of a vertical cliff while being pursued by angry natives and dives into a river 80m below, splashing into the water 25.0m from the base of the cliff. (assume that he leaps perfectly horizontally to the cliffs edge and air resistence is negligible.)

a) How long will it take tarzan to reach the water?

b) How fast must he be able to leap

c) With what velocity will Tarzan hit the water?

Answer 1

a)

80 = 4.9t²

solve for t

b)

vt = 25

solve for v, use t from part a)

c)

velocity in x direction

Vx = v .....from part b)

velocity in y direction

Vy = -9.8t , use t from part a)

total velocity is

V = √(Vx² + Vy²)

at an angle

θ = arctan(Vy/Vx)  

above the horizontal

answers:

a)4.04s

b) 6.19m/s

c) 40.1m/s @ 81.1deg below horizontal

.,.,.,.,

Answer 2

a)  d = Vi(t) + (1/2)(g)t^2

     80 m = 0(t) + (1/2)(9.8 m/s^2)(t^2)

     t = sqrt[2(80m)/9.8 m/s^2]

     t = 4.04 s    ANSWER

b)  Vx = dx/t

     Vx = 25.0 m/4.04 s

     Vx = 6.2 m/s    ANSWER

c)  Vyf = Viy + gt

     Vyf = 0 + (9.8 m/s^2)(4.04 s)

     Vyf = 35.6 m/s

     V = sqrt[Vyf^2 + Vx^2]

     V = sqrt[(35.6 m/s)^2 + (6.2 m/s)^2]

     V = 36.1 m/s

     tanθ = Vyf/Vx

     θ = arctan(Vyf/Vx)

     θ = arctan(35.6/36.1)

     θ = 44.6 deg

     So, V = 36.1 m/s, 44.6 deg below the horizontal    ANSWER

Hope this helps.

teddy boy