Question

(c) A stone is thrown with speed 49 m/s at an angle of 45°
upward with the earth's horizontal surface. Calculate the
horizontal range of the stone (g=9.8 m/s)​

Answer 1

Answer:

range =R = v02 sin 2θ0/g

[49 x 49 ](sin 45 x 2)/ 9.8

49 x 49 / 98 x 10 ^-1

28.5 =range

hope it helps you .. =D

Answer 2

$\huge\sf{\underline{\pink{Answer:-}}}$

Given V = 16m/s, time = t = 4s

let height = h, and horizontal range R

Velocity in upward direction Vy = V sin (30°)= 8m/s,

Velocity in horizontal Vx = V cos(30°) = 8√3

In upward direction,Using,

$s = ut + \frac{1}{2} {at}^{2}$

$h = 8 \times 4 - \frac{1}{2} \times 9.8 \times {4}^{2} = > \: h - 46.4$

hence,Height = 46.4m,

In horizontal direction, Range = Vx × t

$R = 8\sqrt{3} \times 4 = 32\sqrt{3m}$

Hence,Range is 32√3m

hope this will help you...

-3idiots29