Thus, t = 2 * v * sin(theta) / g
For the maximum height, use the kinematic equation: y = v*sin(theta)*t - (1/2)*g*t^2. The time you need to use is half of the total time.
--> y = (v^2)*[sin(theta)]^2/g - (1/2)*g*(v^2)*[sin(theta)]^2 / (g^2)
--> y = (1/2g)*(v^2)*[sin(theta)]^2
To find the horizontal distance, you need the kinematic equation:
x = v*cos(theta)*t + (1/2)*a*t^2, but because there is no acceleration in the x direction, a = 0. Also, v*cos(theta) is the x-component of the initial velocity and t is the total time in the air.
---> x = v*cos(theta)*2*v*sin(theta) / g = [(v^2)/g]*2*sin(theta)*cos(theta)
2sin(theta)cos(theta) = sin(2*theta)
Thus, x = [(v^2)/g]*sin(2*theta)
Now then...since all that is out of the way, you just need to replace some values and you'll have your answers.
A. t of M = 2*v*sin(theta) / (.379g)
B. h of M =[(v^2)*[sin(theta)]^2] / (2 * .379g)
C. d of M =[(v^2)/(.379g)]*sin(2*theta)
Where g is the acceleration due to gravity at Earth's surface (9.81 m/s^2),
v is the initial velocity of the athlete,
theta is the angle the athlete jumps from the ground.
I'm pretty sure that's all correct.