physics problem?

Question

A baseball is thrown at an angle of 25 degrees relative to the ground at a speed of 23.0m/s.If the ball was caught 42.0m from the thrower,how long was it in the air?how high was the tallest spot in the ball's path?

A quarterback throws the football to a stationary receiver who is 31.5m down the field.If the football is thrown at an initial angle of 40.0 degrees to the ground,at what initial speed must the quarterback throw tha ball for it to reach the receiver?What is the ball's highest point during its flight?

Answer 1

1) x(t)=23*cos(25)*t
when x(t)=42
t=2.015 s
y(t)=23*sin(25)*t-.5*g*t^2
the apogee t is 1/2 the total flight time so the highest point is
4.8 m

2) x(t)=v0*cos(40)*t
y(t)=v0*sin(40)*t-.5*g*t^2
when x(t)=31.5 and y(t)=0, we have a catch
31.5=v0*cos(40)*t
.5*g*t^2=v0*sin(40)*t

divide
.5*g*t^2/31.5=tan(40)
solve for t
t=2.32 s
v0=17.71 m/s
The highest point is
6.61 m
j