You hit a golfball vertically upward with your pitching wedge. The position function of the ball is s(t) = -3t^2+30t where the origin is at ground level and the positive direction is vertically upward. (Distance is in metres).

Help!?

Question

A) Find the maximum height reached by the ball
B) Find the velocity of the ball as it reaches the ground
C) Find the acceleration of the ball.

Ezra Adams · Accepted Answer

A) Maximum height is #75 \ m#
B) Velocity at ground level is #-30 \ ms^(-1) #
C) Acceleration is # = -6 \ ms^(-2) # The ball's vertical displacement with respect to the ground is determined by: # s(t) = -3t^2+30t # The rate of change of displacement with respect to time determines the ball's velocity, that is: # v(t) = d/dt s(t) # So if we differentiate #s(t)# wrt #t# we get: # v(t) = -6t+30 # Just prior to the maximum trajectory, the ball's velocity will increase then it will momentarily stop before starting to drop again, ie we seek #v(t)=0# (or, alternatively, given the definition of displacement, we seek a turning point and set the derivative to zero): # v(t) = 0 => -6t+30 = 0 # # :. 6t = 30 # # :. t = 5 # When #t=5#, the displacement is: # s(5) = -3xx25+30xx5 # # " " = -75+150 # # " " = -75+150 # # " " = 75 # So the maximum height is #75 \ m# At ground level we have no displacement, ie #s(t)=0# # s(t) = 0 => -3t(t-10) = 0 # # :. t=0, 10 ' (units) We expected #t=0# as the initial condition, so #t=10# corresponds to the ball reaching ground level after flight. # t = 10 => v(10) = -60+30 = -30 \ ms^(-1)# Since the ball is falling rather than rising, we also anticipate a negative velocity. The rate of change of velocity with respect to time determines the ball's acceleration, or # a(t) = d/dt v(t) = (d^2)/(dt^2) s(t)# So if we differentiate #v(t)# wrt #t# we get: # a(t) = -6 \ ms^(-2) #, which is constant acceleration throughout flight

Axel Alvarez · Answer

undefined To solve the problem, we need to find the time when the golf ball reaches its maximum height and then determine that maximum height.

First, we find the maximum height by finding the vertex of the parabolic function represented by the position function $ s(t) = -3t^2 + 30t $. The vertex of a parabola in the form $ y = ax^2 + bx + c $ is given by the point $ (h, k) $, where $ h = -\frac{b}{2a} $ and $ k = f(h) $. In this case, $ a = -3 $ and $ b = 30 $.

Using the formula $ h = -\frac{b}{2a} $, we find:

$$ h = -\frac{30}{2(-3)} = 5 $$

To find the maximum height, we evaluate the function at $ t = 5 $:

$$ s(5) = -3(5)^2 + 30(5) = -3(25) + 150 = 75 $$

So, the maximum height reached by the golf ball is 75 meters.

Next, to find the time it takes for the ball to reach the maximum height, we use the fact that the maximum height occurs at the vertex of the parabola. Therefore, the time when the ball reaches its maximum height is $ t = 5 $ seconds.

Finally, since the ball is in free-fall motion, we know that it will take the same amount of time to fall back down to the ground as it took to reach its maximum height. Therefore, the total time the ball is in the air is $ 5 + 5 = 10 $ seconds.

You hit a golfball vertically upward with your pitching wedge. The position function of the ball is s(t) = -3t^2+30t where the origin is at ground level and the positive direction is vertically upward. (Distance is in metres). Help!?

A) Find the maximum height reached by the ball B) Find the velocity of the ball as it reaches the ground C) Find the acceleration of the ball.

A) Find the maximum height reached by the ball
B) Find the velocity of the ball as it reaches the ground
C) Find the acceleration of the ball.