Final answer:
To find the velocity of the ball at t=1s, we take the derivative of the equations for x and y with respect to t. The velocity of the ball at t=1s is v=16.24 m/s. To find the horizontal distance the ball travels before hitting the ground, we solve the equation y=0 and find t=2.5s. Plugging this into x=5t gives us a horizontal distance of 12.5 meters.
Step-by-step explanation:
To find the velocity of the ball at t=1s, we need to take the derivative of the equations for x and y with respect to t. The derivative of x=5t is dx/dt=5, which gives us the horizontal velocity of the ball. The derivative of y=2+6t-4.9t^2 is dy/dt=6-9.8t, which gives us the vertical velocity of the ball.
Thus, at t=1s, the vertical velocity of the ball is dy/dt=6-9.8(1)=-3.8 m/s and the horizontal velocity is dx/dt=5 m/s. Therefore, the velocity of the ball at t=1s is v=(dx/dt)^2+(dy/dt)^2=(5)^2+(-3.8)^2=16.24 m/s.
To find the horizontal distance the ball travels before hitting the ground, we need to find the time it takes for the ball to hit the ground. This can be done by setting the equation for y equal to zero and solving for t. The equation for y is y=2+6t-4.9t^2=0. Solving this quadratic equation gives us t=2.5s and t=-0.166s. Since time cannot be negative, we can ignore the negative solution.
The horizontal distance the ball travels before hitting the ground can be found using the equation x=5t. Plugging in t=2.5s, we get x=5(2.5)=12.5m. Therefore, the ball travels a horizontal distance of 12.5 meters before hitting the ground.