Question

A ball throw with a speed of 100ms-1 attain height of 150m(takeg = 98 mg-2)calculate the following
(a)time of flight
(b)angle of projection
(c)range

Answer 1

We can use the following equations to solve this problem:

1. Vertical motion:
v_y = u_y - g*t
h = u_y*t - (1/2)*g*t^2

2. Horizontal motion:
R = u_x*t

3. Initial velocity components:
u_x = u*cos(θ)
u_y = u*sin(θ)

Given:
u = 100 m/s
h = 150 m
g = 9.8 m/s²

(a) Time of flight
First, we need to find the time when the ball reaches the highest point (h = 150 m). We can use the equation for vertical motion:

150 = u_y*t - (1/2)*9.8*t^2

Since we don't know u_y and t yet, we can use the equation for vertical velocity:

v_y = u_y - 9.8*t

At the highest point, the vertical velocity is 0. Therefore, we can write:

u_y = 9.8*t

Plugging this back into the height equation:

150 = (9.8*t)*t - (1/2)*9.8*t^2
150 = (1/2)*9.8*t^2

Now, solve for t:

t^2 = (150*2)/9.8 ≈ 30.6122
t ≈ √30.6122 ≈ 5.53 seconds

Since the ball takes the same amount of time to go up and come back down, the total time of flight is:

T = 2*t ≈ 2*5.53 = 11.06 seconds

(b) Angle of projection
Using the equation for the vertical velocity component:

u_y = u*sin(θ)
u_y = 9.8*t

Plug in the values:

u*sin(θ) = 9.8*5.53

Divide both sides by u:

sin(θ) = (9.8*5.53)/100 ≈ 0.541
θ ≈ arcsin(0.541) ≈ 32.5 degrees

(c) Range
Using the equation for the horizontal motion:

R = u_x*T

We know that u_x = u*cos(θ), so:

R = u*cos(θ)*T

Plug in the values:

R ≈ 100 *cos(32.5) *11.06 ≈ 928.4 meters

The range is approximately 928.4 meters.