A tomato is thrown upward from a bridge 23 m above the ground at 40 m/sec. Take the acceleration of gravity be equal to 9.8 m/s^2

(a) Give formulas for the acceleration, velocity, and height of the tomato at time t.

(b) How high does the tomato go, and when does it reach its highest point?

(c) How long is it in the air?

my equation would be

h = -4.9t^2 + 40t + 23
v = - 9.8t + 40
a = -9.8

b) max h, when v = 0
-9.8t+40=0
t = 40/9 seconds
h= -4.9(40/9)^2 + 40(40/9) + 23 = 103.99 m

c) let's find when it hits the ground, that is, when h = 0
-4.9t^2 + 40t + 23 = 0
4.9t^2 - 40t - 23 = 0
t = (40 ± √2050.8)/9.8
= 8.7 or some negative, which would be silly

It is in the air for appr 8.7 seconds

BTW, that is some feat to throw a tomato about 80 m up into the air !

(a) Formulas for the acceleration, velocity, and height of the tomato at time t are:

Acceleration (a): The acceleration due to gravity, which is always -9.8 m/s^2 since it acts in the opposite direction of motion.
Velocity (v): v = initial velocity + (acceleration * time)
Height (h): h = initial height + (initial velocity * time) + (0.5 * acceleration * time^2)

(b) To find how high the tomato goes, we need to find the maximum height it reaches. At this point, the velocity will be 0 m/s. Using the formula for velocity:

0 = initial velocity + (acceleration * time)

Solving for time:

time = -initial velocity / acceleration
time = -40 m/s / -9.8 m/s^2
time ≈ 4.08 s

To find the maximum height, we can substitute this value of time into the height formula:

h = initial height + (initial velocity * time) + (0.5 * acceleration * time^2)
h = 23 m + (40 m/s * 4.08 s) + (0.5 * -9.8 m/s^2 * (4.08 s)^2)
h ≈ 43.06 m

Therefore, the tomato reaches a height of approximately 43.06 meters.

(c) To determine how long the tomato is in the air, we need to find the total time it takes for the tomato to go up and then come back down. We can find this by doubling the time it took to reach the maximum height:

Total time = 2 * time
Total time ≈ 2 * 4.08 s
Total time ≈ 8.16 s

Therefore, the tomato is in the air for approximately 8.16 seconds.

(a) The formulas for the acceleration, velocity, and height of the tomato at time t can be given as:

Acceleration (a) = -9.8 m/s^2 (due to the acceleration of gravity)
Velocity (v) = initial velocity ( u ) + acceleration (a) * time (t)
Height (h) = initial height (h0) + initial velocity (u) * time (t) + (1/2) * acceleration (a) * time^2

(b) To find out how high the tomato goes and when it reaches its highest point, we can use the formula for the height (h) mentioned in part (a). At the highest point, the velocity of the tomato will be zero. Therefore, we need to find the value of time (t) when the velocity becomes zero.

Given: initial height (h0) = 23 m, initial velocity (u) = 40 m/s, and acceleration (a) = -9.8 m/s^2.

At the highest point, the velocity is zero. So we have:

0 = 40 - 9.8t

Solving for t:

40 = 9.8t
t = 40 / 9.8
t ≈ 4.08 seconds

Substituting the value of t into the height formula:

h = 23 + 40t - (1/2) * 9.8 * t^2
h = 23 + 40 * 4.08 - (1/2) * 9.8 * (4.08)^2

Calculating the height:

h ≈ 23 + 163.2 - 79.871
h ≈ 106.329 m

Therefore, the tomato reaches a maximum height of approximately 106.329 m and it reaches its highest point at approximately 4.08 seconds.

(c) To find out how long the tomato is in the air, we need to determine the total time it takes to reach the ground. Since the tomato is thrown upwards, the total time will be twice the time it takes to reach its highest point (t).

Total time in the air = 2 * t
Total time in the air = 2 * 4.08
Total time in the air ≈ 8.16 seconds

Therefore, the tomato is in the air for approximately 8.16 seconds.

(a) Formulas for the acceleration, velocity, and height of the tomato at time t can be derived by using equations of motion.

1. Acceleration (a):
The tomato is thrown vertically upward, so the acceleration can be considered as the acceleration due to gravity. Therefore, the acceleration (a) will be a constant value of -9.8 m/s^2 (since it acts downward).

2. Velocity (v):
The initial velocity (u) is given as 40 m/s (the tomato is thrown upward). The velocity (v) at any time (t) can be calculated using the formula:
v = u + at

In this case, since the acceleration is acting downward (in the opposite direction to the motion), the formula becomes:
v = u - gt

3. Height (h):
The initial height (h0) is given as 23 m (the bridge height). The height (h) of the tomato at any time (t) can be calculated using the formula:
h = h0 + ut + (1/2)at^2

(b) To find how high the tomato goes and when it reaches its highest point, we need to determine the time at which the velocity becomes zero.

At its highest point, the velocity of the tomato is zero. Therefore, using the formula v = u - gt, we can find the time when v becomes zero:
0 = 40 - 9.8t
t = 40/9.8

To find the height, we can substitute this time (t) into the formula h = h0 + ut + (1/2)at^2:
h = 23 + (40t) + (0.5)(-9.8)(t^2)

(c) To find how long the tomato is in the air, we need to determine the time it takes for the tomato to reach the ground. Since the tomato is thrown upward, it will reach the ground when the height (h) becomes zero.

Using the quadratic equation, we can solve for the time (t) when h = 0 in the formula h = h0 + ut + (1/2)at^2:
0 = 23 + (40t) + (0.5)(-9.8)(t^2)

Solving this equation will give us the time it takes for the tomato to reach the ground, which will give us the duration it is in the air.