The muzzle velocity of a bullet fired from a new rifle is 100 m/s. neglecting air resistance, at the end of one second a bullet fired straight up into the air will have traveled a distance of: . 100mb. 4.9mc. (100+4.9)md. (100-4.9)me. none of these

C. (100+4.9)m

Explanation:

u = 100m/s, t = 1 sec, g = 9.8m/s^2

s = ut + 1/2gt^2

s = 100×1 + 1/2×9.8×1^2 = 100+4.9

s = (100+4.9)m

answer; how far out of place which the object finished relative to the starting location;

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft