A ball is kicked off the ground reaching a maximum height of 60m and lands 80m away. Calculate the initial speed and the angle above the horizontal of the ball when it was kicked

36.87°

Explanation:

Given

Maximum height = 60m

Horizontal distance (range) = 80,m

Required

Initial speed U

Angle of launch

To get the speed, we will use the range formula;

R = U √2H/g

80 = U√2(60)/9.8

80 = U√12.25

80 = 3.5U

U = 80/3.5

U = 22.86m/s

Get the angle of launch

Using the formula

Theta = tan^-1(y/x)

y is the vertical distance

x is the horizontal distance

Theta = tan^-1(60/80)

Theta = tan^-1(0.75)

Theta = 36.87°

Hence the angle of launch is 36.87°

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s