Answer:
it's pray hoped this helped
Answer:
31.75 m/s
Explanation:
h = 41.7 m
Let the initial velocity of the second stone is u
Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.
For first stone:
Use second equation of motion

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7
So, 41.7= 0 + 0.5 x 9.8 x t^2
41.7 = 4.9 t^2
t = 2.92 s ..... (1)
For second stone:
Use second equation of motion

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity
.... (2)
By equation the equation (1) and (2), we get

u = 31.75 m/s
Eh not really sure bout this one
I think 100 mph pushing the car the opposite direction
Answer:
65.87 s
Explanation:
For the first time,
Applying
v² = u²+2as.............. Equation 1
Where v = final velocity, u = initial velocity, a = acceleration, s = distance
From the question,
Given: u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m
Substitute these values into equation 1
v² = 0²+2(1.99)(60)
v² = 238.8
v = √238.8
v = 15.45 m/s
Therefore, time taken for the first 60 m is
t = (v-u)/a............ Equation 2
t = (15.45-0)/1.99
t = 7.77 s
For the final 40 meter,
t = (v-u)/a
Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²
Substitute into the equation above
t = (0-15.45)/-0.266
t = 58.1 seconds
Hence total time taken to cover the distance
T = 7.77+58.1
T = 65.87 s