Height of a baseball - kinematics

[CellCoree]

A major leaguer hits a baseball so that it leaves the bat at a speed of 31.4 m/s and at an angle of 38.1 above the horizontal. You can ignore air resistance

#1: At what two times is the baseball at a height of 10.9s above the point at which it left the bat? Give your answers in ascending order separated with comma.

my work:
x(t) = X(0) + V(0)t + 1/2at^2

10.9 = 0 + 31.4m/s(t) + 1/2(-9.8)t^2
=-4.9t^2 + 31.4t - 10.9

Using both the Quad. Equation and my calculator, i found the same answer. the answer is: $$0.36830144 , 6.0398618$$

can someone help me? i don't know what I am doing wrong

 

[Leong]

Supposed to be 10.9 m in your question.
The velocity of 31.4 m/s is the vector sum of the x and y components. in your equation, you should use the y component of the velocity which is given as 31.4*sin 38.1.

 

[M98Ranger]

.

#2: The height of the baseball can be calculated using the kinematic equation y(t) = y(0) + v(0)sin(θ)t - 1/2gt^2, where y(0) is the initial height (0), v(0) is the initial velocity (31.4 m/s), θ is the angle (38.1°), and g is the acceleration due to gravity (-9.8 m/s^2). Substituting these values into the equation, we get y(t) = 31.4sin(38.1°)t - 4.9t^2. To find the times when the height is 10.9 m, we can set this equation equal to 10.9 and solve for t. This gives us the quadratic equation 4.9t^2 - 31.4sin(38.1°)t + 10.9 = 0. Solving for t using the quadratic formula, we get two possible times: t = 0.3683 s and t = 6.0399 s. These correspond to the times when the ball is at a height of 10.9 m above the point where it left the bat. Therefore, the two times are 0.3683 s and 6.0399 s, in ascending order.