Physics,Physics,Physics?

Question

1. A baseball is hit at 30m/s at an agle of 53 degrees. Immediately, the centerfielder runs at a speed of 4m/s directyl towards home plate and catches the ball. How far was the outfielder from the batter when the ball was hit?
2. A famous motorcyclist plans to jump across a canyon that is .03km wide. To do this, the cyclist plans to jump off of a 30 degree ramp at a speed of 19.6m/s. Will he reach the other side?
WORK PLEASE

Answer 1

1) The equations of motion of the ball are
y(t)=30*sin(53)*t-.5*g*t^2
and
x(t)=30*cos(53)*t

for the outfielder
x(t)=x0-4*t

When the x(t) of the ball=x(t) of the outfielder, and y(t)=0 (assuming the ball is caught at the same height it was hit), we have a catch
solve for x0

x0-4*t=30*cos(53)*t
x0/22=t

0=1.086*x0-0.01*x0^2

x0=108 m


2) is the canyon 0.03km, only 30 m
or 0.3 km, 300 m?

x(t)=19.6*cos(30)*t
To reach the other side t=

1.767 seconds

y(t)=19.6*sin(30)*t-.5*g*t^2

y(t)=2 m, made it

j