Question

A football is thrown from the edge of a cliff from a height of 22 m at a velocity of 18 m/s [39degrees above the horizontal]. A player at the bottom of the cliff is 12 m away from the base of the cliff and runs at a maximum speed of 6.0 m/s to catch the ball.
Is it possible for the player to catch the ball? Support your answer with calculations.

Answer 1

Vo = 18m/s[39o]
Xo = 18*cos39 = 13.99 m/s.
Yo = 18*sin39 = 11.33 m/s.

Y = Yo + g*Tr = 0 @ max ht.
11.33 - 9.8*Tr = 0
9.8Tr = 11.33
Tr = 1.16 s. = Rise time.

hmax = ho + (Y^2-Yo^2)/2g
hmax = 22 + (0-11.33^2)/-19.6 = 28.55 m.
Above gnd.

h = 0.5g*t^2 = 28.55 m.
4.9*t^2 = 28.55
t^2 = 5.83
Tf = 2.41 s = Fall time.


V * (1.16+2.41) = 12
3.57V = 12
V = 3.36 m/s = Required velocity.
Yes.

Answer 2

No. Henry's solution makes sense up until the last bit.

The player is 12m away from the base of the cliff. He's running away from the cliff, or in other words, in the same horizontal direction as the football. Therefore, you need to sub in the time (3.56s) into the equation for the horizontal velocity, to determine the range of the football:

Vh(orXo)=d/t
d=Vh*t
d=13.99*3.5672
d=49.900

Okay, so the player needs to cover a distance of 50m, in order to catch the ball.

V=6.0m/s
t=3.56s
d=?

How far can he travel in 3.56s?
d=vt
d=6*3.56
d=21.4m

21.4m+12m (he was already displaced by this much)

Therefore, the football will travel about 50m, while the player can only run a maximum of 33m. He won't be able to make it.