The springbok, an antelope found in Africa, gets its name from its remarkable jumping ability. When startled, a springbok will leap straight up into the air—a maneuver called a "pronk." A springbok goes into a crouch to perform a pronk. It then extends its legs forcefully, accelerating at 35 m/s2 for 0.70 m as its legs straighten. Legs fully extended, it leaves the ground and rises into the air. How high does it go?

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t (1)

{Vf^{2}-Vo^2}/{2.a} =X (2)

X=Xo+ VoT+0.5at^{2} (3)

X=(Vf+Vo)T/2 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

To solve this problem we must divide it in tow parts, the first is when the springbok accelerates upwards with 35m / S ^ 2, the objective of the first part is to find the final speed using equation number 2

Part 1

Vo=0

a= 35 m/s2

x=0.7m

{Vf^{2}-Vo^2}/{2.a} =X

solving for Vf

[tex]Vf=\sqrt{2ax} =\sqrt{2(35)(0.7)} =7m/s[/tex]

part 2

for part 2, the final velocity of part 1 is the initial velocity of movement 2, and the acceleration is the gravity = -9.81m / s ^ 2.

we use the ecuation number tow(2)

Vo=7m/s

a=g=-9.81m/S^2

Vf=0m/s

{Vf^{2}-Vo^2}/{2.a} =X

{0^{2}-(7)^2}/{2.(-9.81)} =X

X=2.5m

finally the distance traveled is the sum of the distances of part 1 and 2

Xt=2.5+0.7=3.2m