Question

a body is projected obliquely with an initial velocity of 19.6m/s at an angle of 30° with the horizontal.find the maximum height reached and horizontal range​

Answer 1

Answer:

Maximum Height = 4.9 m, Horizontal Range = 33.94 m

Explanation:

In the y-direction :

By conservation of energy:

$E_i$ = $E_f$

Initial KE + Initial PE = Final KE + Final PE

Initial state - When the ball is just thrown, Final State - Ball has reached $H_m_a_x$

$\frac{1}{2}$ m $v^{2}$ + 0 = mgh + 0

$\frac{1}{2}$ m $v^{2}$  = mgh

h = $\frac{v^2}{2g}$ , v = y component of initial velocity = usinθ

$H_m_a_x$ = $\frac{u^2 sin^2 theta}{2g}$ , u = 19.6, θ = 30°

$H_m_a_x$ = 4.9 m

Range R = $u_x$ t

- $u_y$ = -g (time of ascent),  $u_y$ = g (time of descent)

$t_a$ = $u_y$/g, $t_d$ =  $u_y$/g

Total time t = time of ascent + time of descent = 2 $u_y$/g

$u_y$ = ucosθ

t = 2ucosθ/g

$u_x$ = usinθ

R = $u_x$ t

R = 2$u^{2}$sinθcosθ/g

R = $u^{2}$sin(2θ)/g, u = 19.6, θ = 30°

R = 33.94 m

Hope it helps :)