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
