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
Answers
Answer
(a) 30.84m
(b) 5.01s
(c) 136m
(d) 47.7m/s
Explanation
we are given
initial speed, u = 36.6m/s
angle Ф = 42.2°
acceleration of gravity = 9.8m/s²
(a) maximum height = u²sin²Ф/ 2g
= 36.6²sin²(42.2) / 2(9.8)
= 30.84m
(b) total time in air = 2u*sin(Ф)/g
= 2 (36.6)sin(42.2)/9.8
= 5.01s
(c) total horizontal distance covered = u²sin(2Ф)/g
= (36.6)²sin(2*42.2)/9.8
= 136m
(d) the speed of projectile after 1.50s after firing:
To solve this we will split the initial velocity into 2 directions, x and y. Start with finding 'u' in y direction by using the following equation:
u(y) = u*sin(Ф)
= 24.6m/s
now use this equation:
v(y)=u(y)+a*t where v in the final speed
= 24.6 + 9.8(1.50)
= 39.3 m
Ignoring air resistance, assume the velocity in the x direction is constant so it remains Vix which can be solved using:
u(x)=u*cos(Ф)
= 36.6cos(42.2)
= 27.11 m/s
and
v(x)=u(x)
Now, solve for V (final velocity) using:
V= √(vx²)+(vy²)
= √39.3² +27.11²
= 47.7m/s
Answer:
What he said but D = 28.9 m/s ...
Explanation:
... make sure to use a negative acceleration (-9.8m/s^2) as opposed to a positive because gravity is a downward force therefore in this problem it is negative.