after ploting graph for h = 1.1 + 7t - 5t^2 for interval t = 0 to t = 2. i catch a ball again when it is at height at 1.1m given the velocity of ball v in m/s given by v = 7 - 10t, find the speed of the ball when it reaches my hand.
sub t = 1.1 into
v = 7 - 10t
To find the speed of the ball when it reaches your hand, we need to find the derivative of the height function.
The equation for the height function is h = 1.1 + 7t - 5t^2. To find its derivative, we differentiate the equation with respect to time (t).
dh/dt = d/dt (1.1 + 7t - 5t^2)
The derivative of 1.1 with respect to t is 0 since it's a constant.
The derivative of 7t with respect to t is 7 since the derivative of t is 1.
The derivative of -5t^2 with respect to t is -10t since the derivative of t^2 is 2t, and the derivative of -5t^2 is obtained by multiplying 2t with -5.
Therefore, dh/dt = 0 + 7 - 10t = -10t + 7.
Now we're given the velocity function of the ball, v = 7 - 10t. The speed of the ball is the absolute value of the velocity, so we can ignore the negative sign:
speed = |v| = |7 - 10t|
To find the speed of the ball when it reaches your hand, we need to find the time (t) when the ball reaches a height of 1.1m, and then substitute this value of t into the speed equation.
Since we have the equation h = 1.1 + 7t - 5t^2, we can set h equal to 1.1 and solve for t:
1.1 = 1.1 + 7t - 5t^2
Simplifying the equation gives us:
0 = 7t - 5t^2
Rearranging the terms gives us a quadratic equation:
5t^2 - 7t = 0
Factoring out t, we get:
t(5t - 7) = 0
Setting each factor equal to zero:
t = 0 or t = 7/5
We can ignore the solution t = 0 since the time t = 0 corresponds to the initial position of the ball, not when it reaches your hand. Therefore, the ball reaches your hand at t = 7/5.
Substituting this value of t into the speed equation:
speed = |7 - 10(7/5)| = |7 - 14| = |-7| = 7 m/s.
Therefore, the speed of the ball when it reaches your hand is 7 m/s.