Answer:
3.95979 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Here 
Initial velocity of the puck should be 3.95979 m/s
Answer:
A. 148.23 m
B. 2.75 m/s
Explanation:
The following data were obtained from the question:
Time of flight (T) = 11 s
Maximum height (h) =?
Initial velocity (u) =?
Next, we shall determine the time taken for the ball to get to the maximum height. This can be obtained as follow:
Time of flight (T) = 11 s
Time (t) to reach the maximum height =.?
T = 2t
11 = 2t
Divide both side by 2
t = 11/2
t = 5.5 s
NOTE: Time to reach the maximum height is the same as the time taken for the ball to fall back to the plane of projection.
A. Determination of the maximum height to which the ball was thrown.
Time (t) to reach maximum height = 5.5 s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) =?
h = ½gt²
h = ½ × 9.8 × 5.5²
h = 4.9 × 30.25
h = 148.23 m
B. Determination of the initial velocity.
Maximum height (h) reached = 148.23 m
Acceleration due to gravity (g) = 9.8 m/s²
Initial velocity (u) =?
u² = h/2g
u² = 148.23 / (2 × 9.8)
u² = 148.23 / 19.6
Take the square root of both side
u = √(148.23 / 19.6)
u = 2.75 m/s
Answer:
Stopping distance = 40m
Explanation:
Given the following :
Initial speed of vehicle before applying brakes = 72km/hr
Converting km/hr to m/s:
72km/hr = [(72 * 1000)m] / (60 * 60)
72km/hr = 72,000m / 3600s
72km/hr = 20m/s
Deceleration after applying brakes (-a) (negative acceleration) = - 5m/s^2
From the 3rd equation of motion:
v^2 = u^2 + 2as
Where v = final Velocity ; u= Initial Velocity ; a = acceleration and s = distance
Final velocity when the car stops will be 0
Therefore ;
v^2 = u^2 + 2as
0 = 20^2 + 2(-5)(s)
0 = 400 - 10s
10s = 400
s = 400/10
s = 40m
Therefore, the stopping distance of the car = 40 meters
