Answer:
45.89m/s²
Explanation:
Given
Distance S = 305m
Time t = 3.64s
To get the acceleration during this run, we will apply the equation of motion:
S = ut+1/2at²
Substitute the given parameters into the formula and calculate the value of a
305 = 0+1/2 a(3.64)²
304 = 1/2(13.2496)a
304 = 6.6248a
a = 304/6.6248
a = 45.89m/s²
Hence the average acceleration during this run is 45.89m/s²
Answer:
B) t = 1.83 [s]
A) y = 16.51 [m]
Explanation:
To solve this problem we must use the following equation of kinematics.

where:
Vf = final velocity = 0
Vo = initial velocity = 18 [m/s]
g = gravity acceleration = 9.81 [m/s²]
t = time [s]
Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.
A) The maximum height is reached when the final velocity of the ball is zero.
0 = 18 - (9.81*t)
9.81*t = 18
t = 18/9.81
t = 1.83 [s], we found the answer for B.
Now using the following equation.

where:
y = elevation [m]
Yo = initial elevation = 0
y = 18*(1.83) - 0.5*9.81*(1.83)²
y = 16.51 [m]