a circus artist drops a ball from a high wire. The ball takes 1.5 seconds to reach the ground. (i) Find the height of the high wire above the ground (ii) write an expression for the speed of the ball t seconds after it's dropped where 0<t<1.5 (iii) how fast is the ball moving as it hits the ground?
h - 4.9 * 1.5^2 = 0
v = -9.8t
thi
(i) To find the height of the high wire above the ground, we can use the equation for the distance fallen by an object under constant acceleration. The equation is:
s = ut + (1/2)at^2
Where:
s = distance fallen (height of the high wire)
u = initial velocity (0, as the ball is dropped)
a = acceleration due to gravity (approximately -9.8 m/s^2, as it acts downwards)
t = time taken (1.5 seconds)
Plugging in the values:
s = 0(1.5) + (1/2)(-9.8)(1.5)^2
s = (-4.9)(1.5)^2
s = -4.9(2.25)
s = -11.025 meters
Since the height cannot be negative, the height of the high wire above the ground is 11.025 meters.
(ii) To write an expression for the speed of the ball at any time between 0 and 1.5 seconds, we can use the equation for velocity:
v = u + at
Where:
v = final velocity (speed of the ball)
u = initial velocity (0, as the ball is dropped)
a = acceleration due to gravity (approximately -9.8 m/s^2, as it acts downwards)
t = time taken (between 0 and 1.5 seconds)
Plugging in the values:
v = 0 + (-9.8)t
v = -9.8t
So, the expression for the speed of the ball at any time between 0 and 1.5 seconds is -9.8t m/s.
(iii) As the ball hits the ground, its final speed is given by the same expression as in (ii). Plugging in the value of t = 1.5 seconds:
v = -9.8(1.5)
v = -14.7 m/s
Therefore, the ball is moving at a speed of 14.7 m/s as it hits the ground.
To solve this problem, we can use the equations of motion. Let's go step by step to answer each part of the question:
(i) To find the height of the high wire above the ground, we can use the equation for free fall:
h = (1/2) * g * t^2
Where:
h is the height of the high wire above the ground,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time it takes for the ball to reach the ground (given as 1.5 seconds).
Substituting the values into the equation, we get:
h = (1/2) * 9.8 * (1.5)^2
h = 11.025 meters (to three decimal places)
Therefore, the height of the high wire above the ground is 11.025 meters.
(ii) To write an expression for the speed of the ball t seconds after it's dropped, where 0 < t < 1.5, we can use the equation for free fall velocity:
v = g * t
Where:
v is the speed of the ball,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time.
Substituting the values into the equation, we get:
v = 9.8 * t
Therefore, the expression for the speed of the ball t seconds after it's dropped is v = 9.8t (m/s).
(iii) As the ball hits the ground, it would have reached its maximum speed. The speed of the ball just before hitting the ground can be found by substituting t = 1.5 seconds into the expression we derived in part (ii):
v = 9.8 * 1.5
v = 14.7 m/s
Therefore, the ball is moving at a speed of 14.7 m/s just as it hits the ground.