Question

A object is thrown with a velocity of 50 m/s at an angle of 300 with horizontal calculate 1)Range

2) Time of flight 3) Maximum height reached.​

Answer 1

Given: Initial Velocity, u = 50m/s

Acceleration due to gravity, $g = 9.8m/s^{2}$

Angle, $\theta = 30^{0}$

Find: We have to calculate the (i) Range, (ii) Time of Flight, (iii) Maximum height

Solution:

(i) We have the expression to calculate the Range,

$R = \dfrac{u^{2}\sin2\theta }{g}$

Now we put the value and get the range,

$R = \dfrac{50^{2}\sin60^{0} }{9.8}\\\\R = \dfrac{2165.0635}{9.8}\\\\R = 220.9248m$

Here the range is 220.9248m

(ii) We have the expression to calculate the Time of Flight,

$T = \dfrac{2u sin\theta}{g}$

Now we put the value and get the value of time of the flight,

$T = \dfrac{2\times50\times sin30 }{9.8}\\\\T= \dfrac{50}{9.8}\\\\T= 5.1020s$

Here the time of the flight is 5.1020s

(iii) We have the expression to calculate the maximum height reached,

$H = \dfrac{u^{2}(sin\theta)^{2} }{2g}$

Now we put the value and get the value of the maximum height reached,

$H = \dfrac{50^{2}\times (sin30)^{2} }{2\times 9.8} \\\\H = \dfrac{2500\times 0.25}{19.6}\\\\H = \dfrac{625}{19.6}\\\\H = 31.8877m$

Here the maximum height reached is 31.8877m