Question

nutso the clown does a human canon ball routine. He wants the peak of his flight to be 10m up and 20m out horizontally from the launch site (through a flaming hoop). Use both unit vector and magnitude, direction notations,

A) What should his launch velocity be?

B) What will his postion be at T=1s

C) What will his height be when he is 10m out from the launch site?

D) For the same launch speed, what is his maximum range?

PLEASE BE SPECIFIC

Answer 1

Hmax = (v0y^2) /2 g

v0y = sqrt(2 x 9.8 x 10) = 14 m/s

time of flight, yf - yi = v0y t + ay t^2 /2

0 -0 = 14 t - 9.8 t^2 / 2

t = 2.86 sec

so R = v0x t

v0x = (20/2.86) = 7 m/s

(A) launch velocity = 7 i + 14 j m/s

magnitude = sqrt(7^2 + 14^2) = 15.6 m/s

direction = tan^-1(14/7) =63 deg  

(B) x = 7 x 1 = 7 m

y = (14 x 1) + (-9.8 x 1^2 / 2) = 9.1 m

position = (7, 9.1)

distance = 11.5 m

direction = 52.4 deg

(C) t = 10/7 = 1.43 sec

h = (14 x 1.43) + (-9.8 x 1/43^2 / 2)

h = 10 m

(D) Rmax = v^2 / g = 15.6^2 / 9.8

= 24.8 m