Question

Q4. A grasshopper jumps at an angle of

30° to the horizontal with a take off
speed of 3m/s
1.
What is the height of its jump? (2)
II.
How long is it above the ground?
(2)
III.
What is the range of its jump? (2)​

Answer 1

i. The height of its jump is approximately 0.115 m

ii. The time of flight of its jump is approximately 0.306 seconds

iii. The range of its jump is approximately 0.795 m

Step-by-step explanation:

The angle at which the grasshopper jumps, θ = 30°

The speed with which the grasshopper takes off, u = 3 m/s

i. The height of its jump 'h', is given by the following relation;

`h = (u^2 * sin^2 \theta)/(2 * g)`

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

`h \approx (3^2 * sin^2 (30^(\circ)))/(2 * 9.81) = (25)/(218) \approx 0.115`

The height of its jump, h ≈ 0.115 m

ii. The time of flight of its jump, 't', is given as follows;

`The \ time \ of \ flight, \, t = (2 * u * sin \theta)/( g)`

Therefore;

`t \approx \frac{2 * 3 * sin (30 ^ {\circ})}{ 9.81} = (100)/(327) \approx 0.306`

The time of flight of its jump, t ≈ 0.306 seconds

iii. The range of the jump is given by the following projectile motion equation for the range as follows;

`R = (u^2 * sin (2 * \theta))/( g)`

Therefore;

`R \approx \frac{3^2 * sin (2 * 30^ {\circ})}{ 9.81} = \frac{41659} {52433} \approx 0.795`

The range of the jump, R ≈ 0.795 m.