A golfer hits a golf ball at an angle of 35° above horizontal. If the initial velocity is 55 m/s, how long does the ball soar before it hits the ground?
initial speed up = 55 sin 35 = Vi
v = Vi - 9.8 t
v = 0 at top
so
0 = 55sin 35 - 9.8 t
solve for t
6.54
sorry 7.85
First statement is correct, bottom two answers are wrong. Remember to multiply time by 2 to give the time up and time down again.
To find the time it takes for the golf ball to hit the ground, we need to break down the initial velocity into its horizontal and vertical components.
The horizontal component of the initial velocity can be calculated using the formula:
Vx = V * cos(θ)
where Vx is the horizontal velocity component, V is the initial velocity (55 m/s), and θ is the angle of launch (35°). Plugging in the values:
Vx = 55 * cos(35°)
The vertical component of the initial velocity can be calculated using the formula:
Vy = V * sin(θ)
where Vy is the vertical velocity component. Plugging in the values:
Vy = 55 * sin(35°)
Now that we have the vertical component of the initial velocity, we can find the time taken for the ball to hit the ground using the formula:
t = 2 * Vy / g
where g is the acceleration due to gravity (9.8 m/s²). Plugging in the values:
t = 2 * (55 * sin(35°)) / 9.8
Simplifying this equation will give us the time taken for the ball to hit the ground.