Question

Tx= vo2/g Sin20 (0 with line through the middle) where Tx= horizontal range, vo= initial velocity, o firing angle at fixed velocity. At which value of 0, r, will attain its maximum value?

Please write clearly and neatly, showing appropriate work. No credit will be given for ambiguous work Question 1: (a) How does the range of the projectile depends of angle of projection? Discuss the answer with formula? d neatly, showing appropriate work. No credit will be given for illegible or (b) Given Formula R-=-sin29 Where Rx = Horizontal Range, vo-Initial Velocity, e-Firing angle At fixed initial velocity at what value of e, R, will attain its maximum value? Question 2: A projectile was fired from atop an elevation at various angles from horizontal surfice The distance was measured for multiple firing angles of each trial [Table 11. Determine the initial velocity of the projectile (Given, g -9.8m/s', Absolute value A,-58 m/s) 40 Points Table: 1 Trial 1 Firing Angle ( Answer must include but not hited to- Horizontal Distance (cm□ 20 30 40 45 50 60 70 259.1 296.3 338.5 337.0 332.2 286.3 a) Theory b) Calculation c)Erro e) Discussion 2 4 6 d) Result 215.2

Answer 1

1.

Range of projectile fired with initial angle with horizontal is defined as the horizontal distance travelled by projectile. Range of projectile is given by R_x = v^2_theta * sin 2 theta/g here, theta is angle of projection. This relation shows us that range of projectile depends on angle of projectile. 2. R will attain maximum value at the maximum value of sin 2 theta maximum value of sin 2 theta = 1 i e 2 theta = 90 degrees hence theta = 45 degrees Hence, Range of projectile is maximum if the projectile is fired at an angle of 45 degrees w.r.t horizontal. 3.

Range of projectile fired with initial angle with horizontal is defined as the horizontal distance travelled by projectile.

Range of projectile is given by

$R_{x} =\frac{v_{0}^{2}* sin2\theta }{g}$

Here, $\theta$ is angle of projection. This relation shows us that range of projectile depends on angle of projectile.

2. R will attain maximum value at the maximum value of $sin2\theta$ .

Maximum value of $sin2\theta$ = 1

i.e, $2\theta = 90 degrees$

hence theta = 45 degrees

Hence, Range of projectile is maximum if the projectile is fired at an angle of 45 degrees w.r.t horizontal.

3.