Question

A projectile is fired with an initial speed of 36.6 m/s at an angle of 42.2° above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered (that is, the range), and (d) the speed of the projectile 1.50 s after firing.

Answer 1

Step-by-step explanation:

a) Maximum height id expressed as;

H = u²sin² theta/2g

H = 36.6²(sin42.2)²/2(9.8)

H = 1,339.56(0.6717)/19.6

H = 899.81/19.6

H = 45.91m

Hence the maximum height is 45.91m

b) The Time of flight is the total time in air expressed as;

T = 2usin theta/g

T = 2(36.6)sin42.2/9.8

T = 73.2sin42.2/9.8

T = 49.17/9.8

T = 5.017secs

Hence the total time in air is 5.017scs

c) Range = U²sin2(theta)/g

Range = 36.6²sin2(42.2)/9.8

Range = 1,339.5(0.9952)/9.8

Range = 1,333.11/9.8

Range = 136.03m

Hence the range is 136.03m

d) USing the rime of flight formula;

T =2Usintheta/g

1.5 = 2Usin42.2/9.8

2Usin42.2 =1.5*9.8

2Usin42.2 = 14.7

U = 14.7/2sin42.2

U = 14.7/1.3434

U = 10.94m/s

Hence the speed of the projectile is 10.94m/s