Question

A grasshopper hops along a level road. on each hop, the grasshopper launches itself itself at an angle of 45 degrees. and achieves a range of 0.77 m. what is the average horizontal speed of the grasshopper as it hops along the road? assume that the time spent on the ground between hops is negligible.

Answer 1

Below is the solution:

 R = v^2 sn(2θ) / g 
v = [R g / sin(2θ)]^1/2 
vx = v cosθ = cosθ [R g / sin(2θ)]^1/2 
θ = π/4 
then 
vx = cos(π/4) [R g / sin(π/2)]^1/2 = cos(π/4) (R g)^1/2 
vx = cos45*(0.82*9.8)^1/2
vx = 2 m/s

Answer 2

Answer: 1.9 m/s

Explanation:

1) Data:

height, y = 0

range, x = 0.77m

angle, α = 45°

question, horizontal velocity, Vox = ?

2) Physical principles and formulas

Projectile motion.

$y=V_{0y}t-gt^{2} /2\\ \\ x=V_{0x}t\\ \\ V_{0y}=V_0sin\alpha \\ \\ V_{0x}=V_0cos\alpha$

3) Solution

$0.77=V_0cos(45)t\\ t=0.77/(V_0cos(45))$

$y=0=V_0sin(45)t-gt^{2} /2$

Factor

$0=t[V_0sin(45)-gt/2]$

Discard t = 0 and solve for the other factor:

$gt/2=V_0sin(45)\\ \\ V_0= gt/(2sin(45))=9.81[0.77/V_{0}cos(45)]/[2sin(45)]\\ \\ V_0^2=9.81[0.77/(2cos(45)sin(45)=9.81[0.77/sin(90)]=7.55\\ \\ V_0}=\sqrt{7.55} =2.8$

Find Vx:

$Vx=V_{0x}=V_{0}cos(45)=2.75(\sqrt{2} /2]=1.9$

Vx = 1.9m/s ← answer