Answer:
The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.
Explanation:
Given that,
Initial velocity u= 128 ft/sec
Equation of height
....(I)
(a). We need to calculate the maximum height
Firstly we need to calculate the time

From equation (I)




Now, for maximum height
Put the value of t in equation (I)


(b). The number of seconds it takes the object to hit the ground.
We know that, when the object reaches ground the height becomes zero




Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.
The answer I'm going with is false
Solution :-
Given :
Distance 1 = 30 km
Distance 2 = 70 km
We know that speed = distance/time
and, Average speed = total distance/total time taken
When the train acquired a speed of 30 km/hr, the time taken = 30/30 = 1 hour
Average speed = 9distance 1 + distance 2)/(time 1 + time 2)
AS time 2 or t2 is time taken for the second part of the journey of 70 km
⇒ 40 = 100/(1 + t2)
⇒ 40 + 40t2 = 100
⇒ 40t2 = 100 - 40
⇒ 40t2 = 60
⇒ t2 = 60/40
⇒ t2 = 1.5
So, t2 or time taken to travel the second part of the journey is 1.5 hours.
Speed of the second part of the journey = distance 2/time 2
⇒ 70/1.5
⇒ 46.666 km/hr or 46.7 km/hr.
Hence the answer is = 46.666 km/hr or 46.7 km/hr.
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Answer:
Ax = 0
Ay = 6 m
Bx = 8 cos phi = cos 34 = 6.63 m
By = 8 sin phi = 8 sin (-34) = -4.47 m
Rx = Ax + Bx = 0 + 6.63 = 6.63 m
Ry = Ay + By = 6 - 4.47 = 1.53 m
R = (6.63^2 + 1.53^2)^1/2 = 6.80 m
tan theta = Ry / Rx = 1.53 / 6.8 = ,225
theta = 12.7 deg