A ball is thrown vertically upward from the ground with an initial velocity of sixty four feet per second. if the positive direction is up to 't' seconds is the time the has elapse since the ball was thrown and 's' is the distance of the ball from the starting point at 't' seconds. Then, the equation is s is equal to negative sixteen plus sixty four t.
Find the instantaneous velocity of the ball when it reaches the ground.
no
s = -16 t^2 + 64 t
If the ball left the ground with velocity + 64 ft/x then it returns to earth with velocity -64 ft/s.
Your parabola is symmetrical and the vertex is at the top,
To find the instantaneous velocity of the ball when it reaches the ground, we need to determine the time it takes for the ball to reach the ground.
In the given situation, the ball is thrown vertically upward, and we know that the positive direction is up. The equation given is s = -16 + 64t, where 's' is the distance of the ball from the starting point at time 't'.
To find the time it takes for the ball to reach the ground, we need to find when s = 0. By substituting s = 0 in the equation, we get:
0 = -16 + 64t
By simplifying this equation, we get:
16 = 64t
To solve for 't', we divide both sides of the equation by 64:
t = 16/64
t = 0.25 seconds
Therefore, it takes 0.25 seconds for the ball to reach the ground.
To find the instantaneous velocity of the ball at this time, we need to take the derivative of the equation s with respect to time t.
s = -16 + 64t
Taking the derivative of s with respect to t, we get:
ds/dt = 64
The derivative of s with respect to t is ds/dt, which represents the instantaneous velocity of the ball at time t.
So, the instantaneous velocity of the ball when it reaches the ground is 64 feet per second.