
Actually Welcome to the Concept of the Projectile Motion.
Since, here given that, vertical velocity= 50m/s
we know that u*sin(theta) = vertical velocity
so the time taken to reach the maximum height or the time of Ascent is equal to
T = Usin(theta) ÷ g, here g = 9.8 m/s^2
so we get as,
T = 50/9.8
T = 5.10 seconds
thus the time taken to reach max height is 5.10 seconds.
According to Ohm's law for a portion of the circuit we have:
U=RI=>I=U/R=24/3=8 A
The correct answer is B
8/4 = y/y-x
8y - 8x = 4y
y = 2x
y = 2 x 4
y = 8
Hope this helps
Explanation:
The given data is as follows.
Spring constant (k) = 78 N/m, 
Mass of block (m) = 0.50 kg
According to the formula of energy conservation,
mgh sin
h =
= 
= 0.64 m
Thus, we can conclude that the distance traveled by the block is 0.64 m.