A ball is thrown vertically upwards at 3 m s-1 from 1m above ground level
(I) write down the equation for the height of the ball above the ground after t seconds (while the ball is in the air).
(II) use your answer to find the time it takes the ball to reach the ground.
(iii) how fast is the ball moving just before it hits the ground
(IV)State an assumption you have made
34
(I) The equation for the height of the ball above the ground after t seconds is given by:
h(t) = h₀ + v₀t - (1/2)gt²
where:
h(t) = height of the ball above the ground after t seconds
h₀ = initial height (1m above ground level in this case)
v₀ = initial velocity (3 m/s upwards in this case)
g = acceleration due to gravity (assumed to be 9.8 m/s²)
(II) To find the time it takes for the ball to reach the ground, we need to find the value of t when the height of the ball becomes 0.
Setting h(t) = 0:
0 = 1 + (3)t - (1/2)(9.8)t²
Simplifying the equation:
4.9t² - 3t - 1 = 0
Using the quadratic formula, we find:
t = (-b ± √(b² - 4ac)) / (2a)
where a = 4.9, b = -3, and c = -1.
Solving for t, we get:
t ≈ 0.42 seconds or t ≈ 0.10 seconds (taking the positive value as time cannot be negative)
Therefore, it takes approximately 0.42 seconds for the ball to reach the ground.
(iii) To find the speed of the ball just before it hits the ground, we need to find the velocity at that time.
Using the equation for velocity:
v(t) = v₀ - gt
Substituting the values, we get:
v(t) = 3 - 9.8(0.42)
v(t) ≈ 3 - 4.116
v(t) ≈ -1.116 m/s
Therefore, the ball is moving at approximately -1.116 m/s just before it hits the ground. The negative sign indicates that the velocity is in the opposite direction of the initial velocity.
(IV) An assumption made in this scenario is that there is no air resistance acting on the ball during its motion.
(I) To write down the equation for the height of the ball above the ground after t seconds, we need to consider the motion of the ball. The ball is thrown vertically upwards, so its initial velocity is positive (3 m/s) and its acceleration is negative (due to gravity, which is approximately -9.8 m/s^2).
Using the kinematic equation for vertical motion:
h = h0 + v0t + (1/2)at^2
Where:
h = height of the ball above the ground after t seconds
h0 = initial height of the ball above the ground (1m)
v0 = initial velocity of the ball (3 m/s)
a = acceleration (−9.8 m/s^2, negative because it opposes the initial velocity)
t = time (in seconds)
Plugging in the values, the equation for the height of the ball above the ground after t seconds is:
h = 1 + 3t - 4.9t^2
(II) To find the time it takes for the ball to reach the ground, we need to set the height h to zero and solve for t. This will give us the time when the ball reaches the ground.
0 = 1 + 3t - 4.9t^2
Rearranging the equation and setting it equal to zero:
4.9t^2 - 3t - 1 = 0
Now, we can solve this quadratic equation to find the value(s) of t.
(iii) To find how fast the ball is moving just before it hits the ground, we need to calculate the velocity (v) by taking the derivative of the height equation with respect to time (t):
v = dh/dt = d(1 + 3t - 4.9t^2)/dt
Differentiating each term with respect to t:
v = 3 - 9.8t
Substituting the value of t at the time of impact (which we found in part II), we can then calculate the velocity.
(IV) One assumption made in this scenario is that there is no air resistance acting on the ball. Realistically, air resistance would affect the motion of the ball and would need to be considered for a more accurate calculation.