Question

a man standing at a distance of 80m from a house throws a ball which just enters a window 40m high if g 10m/s square the angle and velocity of projection are​

Answer 1

Answer: 45 ° and velocity will be 40m/s

Explanation: s=ut+1/2at square and for angle of projection Q=tan^-1(4/n)

Answer 2

Explanation:

Given a man standing at a distance of 80 m from a house throws a ball which just enters a window 40 m high if g 10 m/s square the angle and velocity of projection are

• A man from a house is at a distance of 80 m, he throws the ball and it enters the window that is 40 m high.
• We need to find the angle and velocity of projection.
•                                   h = u^2 sin^2 theta / 2g = 40 ---------- 1
•                                    R = u^2 sin 2 theta / g = 2 x 80 -------- 2
•                            Dividing 1 and 2 we get
•                            u^2 sin^2 theta / 2 g x g / u^2 sin 2 theta = 40 / 2 x 80
•                                                                 tan theta / 4 = ¼
•                                                                 tan theta = 1
•                                                                  Or theta = 45 degree
•                                     Now R = u^2 sin 2 theta / g
•                                            80 x 2 = u^2 sin 2 x 45 / 10
•                                             160 = u^2 sin 90 / 10
•                                             160 = u^2 x 1 / 10
•                                              So u^2 = 1600
•                                               Or u = 40 m/s

Reference link will be

https://brainly.in/question/1468067