Question

If the range of the projectile is 4.3 m, the time-of-flight is T = 0.829 s, and air resistance is negligible, determine the following.a. What is the launch angle of the projectile?b. What is the initial speed of the projectile?c. Express the maximum height reached by a projectile like the one in this problem but with any range and time of flight in terms of g and T.

Answer 1

Answer

given,

range of the projectile = 4.3 m

time of flight = T = 0.829 s

$v =\dfrac{d}{T}$

$v =\dfrac{4.3}{0.829}$

     v = 5.19 m/s

vertical component of velocity of projectile

v_y = gt'

$v_y = 9.8 \times {\dfrac{T}{2}}$

$v_y = 9.8 \times {\dfrac{0.829}{2}}$

$v_y =4.06\ m/s$

a) Launch angle

 $\theta = tan^{-1}(\dfrac{v_y}{v})$

 $\theta = tan^{-1}(\dfrac{4.06}{5.19})$

    θ = 38°

b) initial speed of projectile

  $v = \sqrt{v_x^2 + v_y^2}$

  $v = \sqrt{5.19^2 + 4.06^2}$

         v = 6.59 m/s

c) maximum height reached by the projectile

     $y_{max}=v_{avg}t'$

     $y_{max}=\dfrac{1}{2}v_y\dfrac{T}{2}$

     $y_{max}=\dfrac{1}{2}\times(g\dfrac{T}{2})\times\dfrac{T}{2}$

     $y_{max}=\dfrac{gT^2}{8}$