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;
[tex]h = \dfrac{u^2 \times sin^2 \theta}{2 \times g}[/tex]
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
Therefore;
[tex]h \approx \dfrac{3^2 \times sin^2 (30^{\circ})}{2 \times 9.81} = \dfrac{25}{218} \approx 0.115[/tex]
The height of its jump, h ≈ 0.115 m
ii. The time of flight of its jump, 't', is given as follows;
[tex]The \ time \ of \ flight, \, t = \dfrac{2 \times u \times sin \theta}{ g}[/tex]
Therefore;
[tex]t \approx \dfrac{2 \times 3 \times sin (30 ^ {\circ})}{ 9.81} = \dfrac{100}{327} \approx 0.306[/tex]
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;
[tex]R = \dfrac{u^2 \times sin (2 \times \theta)}{ g}[/tex]
Therefore;
[tex]R \approx \dfrac{3^2 \times sin (2 \times 30^ {\circ})}{ 9.81} = \dfrac{41659} {52433} \approx 0.795[/tex]
The range of the jump, R ≈ 0.795 m.
