Question

Q4. A grasshopper jumps at an angle of

Answer 1

Answer:

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

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;

$h = \dfrac{u^2 \times sin^2 \theta}{2 \times g}$

Where;

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

Therefore;

$h \approx \dfrac{3^2 \times sin^2 (30^{\circ})}{2 \times 9.81} = \dfrac{25}{218} \approx 0.115$

The height of its jump, h ≈ 0.115 m

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

$The \ time \ of \ flight, \, t = \dfrac{2 \times u \times sin \theta}{ g}$

Therefore;

$t \approx \dfrac{2 \times 3 \times sin (30 ^ {\circ})}{ 9.81} = \dfrac{100}{327} \approx 0.306$

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;

$R = \dfrac{u^2 \times sin (2 \times \theta)}{ g}$

Therefore;

$R \approx \dfrac{3^2 \times sin (2 \times 30^ {\circ})}{ 9.81} = \dfrac{41659} {52433} \approx 0.795$

The range of the jump, R ≈ 0.795 m.