Question

A projectile is fired with an initial speed of 60.3 m/s at an angle of 34.2 above the horizontal on a long flat firing range.

Answer 1

Answer:

A.) H = 58.6 m

B.) T = 6.92 s

C.) 345.12 m

D.) V = 22.13 m/s

E.) Ø = 32.1 degree

Explanation:

Given that the

initial speed U = 60.3 m/s

Angle Ø = 34.2 degree

A.) At maximum height, final velocity V is equal to zero.

Using the third equation of motion under gravity.

V^2 = U sin Ø^2 - 2gH

Substitute for U and g. Where g = 9.8 m/s^2

0 = (60.3 sin 34.2)^2 - 2 × 9.8 × H

1148.78 = 19.6 H

H = 1148.78/19.6

H = 58.6 m

B.) To Determine the total time in the air, let us use the formula

V = UsinØ - gt

At maximum height, V = 0

t = UsinØ/g

Total time T = 2t

Therefore, T = 2UsinØ/g

T = (2 × 60.3 × sin 34.2)/9.8

T = 67.79/9.8

T = 6.92 s

C.) To determine the total horizontal distance covered which is the range, we will use second equation of motion.

S = UcosØT - 1/2gt^2

Where S = range R

g = 0, since the range is not a vertical distance

T = total time

Substitute all the parameters into the formula

R = 60.3 cos 34.2 × 6.92

R = 345.12 m

D.) After 1.2 s firing,

V = UsinØ - gt

Where t = 1.2 s

Substitute into the formula

V = 60.3 × sin34.2 - 9.8 × 1.2

V = 33.89 - 11.76

V = 22.13 m/s

Therefore the speed of the projectile 1.20 s after firing is 22.13 m/s

E.) The direction will be determined by using the formula

t = VsinØ/ g

Cross multiply

VsinØ = gt

Make SinØ the subject of formula

SinØ = gt/V

SinØ = (9.8×1.2)/22.13

Sin Ø = 11.76/22.13

Sin Ø = 0.53

Ø = sin^-1( 0.53 )

Ø = 32.1 degree